Ahlfors sol1
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About This Presentation
importent book sol
Slide Content
Solutions toAhlfors' ComplexAnalysis
By:
DustinSmith
By:
DustinSmith
Contents
1 Complex Numbers 5
1.1The Algebra of Complex Numbers 5
1.1.1Arithmetic Operations 5
1.1.2Square Roots 6
1.1.3Justication 7
1.1.4Conjugation, Absolute Value 7
1.1.5Inequalities 10
1.2The Geometric Representation of Complex Numbers 11
1.2.1Geometric Addition and Multiplication 11
1.2.2The Binomial Equation 12
1.2.3Analytic Geometry 14
1.2.4The Spherical Representation 15
2 Complex Functions 17
2.1Introduction to the Concept of Analytical Function 17
2.1.1Limits and Continuity 17
2.1.2Analytic Functions 17
2.1.3Polynomials 20
2.1.4Rational Functions 20
2.2Elementary Theory of Power Series 22
2.2.1Sequences 22
2.2.2Series 22
2.2.3Uniform Convergence 22
2.2.4Power Series 24
2.2.5Abel's Limit Theorem 27
2.3The Exponential and Trigonometric Functions 27
2.3.1The Exponential 27
2.3.2The Trigonometric Functions 27
2.3.3Periodicity 29
2.3.4The Logarithm 29
3 Analytic Functions as Mappings
3.1Elementary Point Set Topology 33
3.1.1Sets and Elements 33
3.1.2Metric Spaces 33
3.1.3Connectedness 34
3.1.4Compactness 36
3.1.5Continuous Functions 37
3.1.6Topological Spaces 37
3.2Conformality 37
3
1 Complex Numbers 5
1.1The Algebra of Complex Numbers 5
1.1.1Arithmetic Operations 5
1.1.2Square Roots 6
1.1.3Justication 7
1.1.4Conjugation, Absolute Value 7
1.1.5Inequalities 10
1.2The Geometric Representation of Complex Numbers 11
1.2.1Geometric Addition and Multiplication 11
1.2.2The Binomial Equation 12
1.2.3Analytic Geometry 14
1.2.4The Spherical Representation 15
2 Complex Functions 17
2.1Introduction to the Concept of Analytical Function 17
2.1.1Limits and Continuity 17
2.1.2Analytic Functions 17
2.1.3Polynomials 20
2.1.4Rational Functions 20
2.2Elementary Theory of Power Series 22
2.2.1Sequences 22
2.2.2Series 22
2.2.3Uniform Convergence 22
2.2.4Power Series 24
2.2.5Abel's Limit Theorem 27
2.3The Exponential and Trigonometric Functions 27
2.3.1The Exponential 27
2.3.2The Trigonometric Functions 27
2.3.3Periodicity 29
2.3.4The Logarithm 29
3 Analytic Functions as Mappings
3.1Elementary Point Set Topology 33
3.1.1Sets and Elements 33
3.1.2Metric Spaces 33
3.1.3Connectedness 34
3.1.4Compactness 36
3.1.5Continuous Functions 37
3.1.6Topological Spaces 37
3.2Conformality 37
3
1 Complex Numbers
1.1 The Algebra of Complex Numbers
1.1.1 Arithmetic Operations
1.
(1+2i)
3
,
5
-3+4i
,
2+i
3-2i
2
, (1+i)
n
+ (1-i)
n
For the rst problem, we have(1+2i)
3
= (-3+4i)(1+2i) = -11-2i . For the second problem, we
should multiple by the conjugate ¯z= -3-4i.
5
-3+4i
-3-4i
-3-4i
=
-15-20i
25
=
-3
5
-
4
5
i
For the third problem, we should rst multiple by ¯z=3+2i.
2+i
3-2i
3+2i
3+2i
=
8+i
13
Now we need to just square the result.
1
169
(8+i)
2
=
63+16i
169
For the last problem, we will need to nd the polar form of the complex numbers. Letz1=1+i
andz2=1-i . Then the modulus ofz1=
p
2=z2
. Let1and2be the angles associated withz1
andz2, respectively. Then1=arctan(1) =
4
and2=arctan(-1) =
-
4
. Thenz1=
p
2e
i=4
and
z2=
p
2e
-i=4
.
z
n
1
+z
n
2
=2
n=2
e
ni=4
+e
-ni=4
=2
n=2+1
e
ni=4
+e
-ni=4
2
=2
n=2+1
cos
n
4
2.z=x+iy(xandyreal), nd the real and imaginary parts of
z
4
,
1
z
,
z-1
z+1
,
1
z
2
Forz
4
, we can use the binomial theorem since(a+b)
n
=
P
n
k=0
n
k
a
n
b
n-k
. Therefore,
(x+iy)
4
=
4
0
(iy)
4
+
4
1
x(iy)
3
+
4
2
x
2
(iy)
2
+
4
3
x
3
(iy) +
4
4
x
4
=y
4
-4xy
3
i-6x
2
y
2
+4x
3
yi+x
4
Then the real and imaginary parts are
u(x,y) =x
4
+y
4
-6x
2
y
2
v(x,y) =4x
3
y-4xy
3
For second problem, we need to multiple by the conjugate ¯z.
1
x+iy
x-iy
x-iy
=
x-iy
x
2
+y
2
5
1.1 The Algebra of Complex Numbers
1.1.1 Arithmetic Operations
1.
(1+2i)
3
,
5
-3+4i
,
2+i
3-2i
2
, (1+i)
n
+ (1-i)
n
For the rst problem, we have(1+2i)
3
= (-3+4i)(1+2i) = -11-2i . For the second problem, we
should multiple by the conjugate ¯z= -3-4i.
5
-3+4i
-3-4i
-3-4i
=
-15-20i
25
=
-3
5
-
4
5
i
For the third problem, we should rst multiple by ¯z=3+2i.
2+i
3-2i
3+2i
3+2i
=
8+i
13
Now we need to just square the result.
1
169
(8+i)
2
=
63+16i
169
For the last problem, we will need to nd the polar form of the complex numbers. Letz1=1+i
andz2=1-i . Then the modulus ofz1=
p
2=z2
. Let1and2be the angles associated withz1
andz2, respectively. Then1=arctan(1) =
4
and2=arctan(-1) =
-
4
. Thenz1=
p
2e
i=4
and
z2=
p
2e
-i=4
.
z
n
1
+z
n
2
=2
n=2
e
ni=4
+e
-ni=4
=2
n=2+1
e
ni=4
+e
-ni=4
2
=2
n=2+1
cos
n
4
2.z=x+iy(xandyreal), nd the real and imaginary parts of
z
4
,
1
z
,
z-1
z+1
,
1
z
2
Forz
4
, we can use the binomial theorem since(a+b)
n
=
P
n
k=0
n
k
a
n
b
n-k
. Therefore,
(x+iy)
4
=
4
0
(iy)
4
+
4
1
x(iy)
3
+
4
2
x
2
(iy)
2
+
4
3
x
3
(iy) +
4
4
x
4
=y
4
-4xy
3
i-6x
2
y
2
+4x
3
yi+x
4
Then the real and imaginary parts are
u(x,y) =x
4
+y
4
-6x
2
y
2
v(x,y) =4x
3
y-4xy
3
For second problem, we need to multiple by the conjugate ¯z.
1
x+iy
x-iy
x-iy
=
x-iy
x
2
+y
2
5
so the real and imaginary parts are
u(x,y) =
x
x
2
+y
2
v(x,y) =
-y
x
2
+y
2
For the third problem, we have
x-1+iy
x+1-iy
. Then ¯z=x+1+iy.
x-1+iy
x+1-iy
x+1+iy
x+1+iy
=
x
2
-1+2xyi
(x+1)
2
+y
2
Then real and imaginary parts are
u(x,y) =
x
2
-1
(x+1)
2
+y
2
v(x,y) =
2xy
(x+1)
2
+y
2
For the last problem, we have
1
z
2
=
x
2
-y
2
-2xyi
x
4
+2x
2
y
2
+y
4
so the real and imaginary parts are
u(x,y) =
x
2
-y
2
x
4
+2x
2
y
2
+y
4
v(x,y) =
-2xy
x
4
+2x
2
y
2
+y
4
3.
-1i
p
3
2
3
=1and
1i
p
3
2
6
=1.
Both problems will can be handled easily by converting to polar form. Letz1=
-1i
p
3
2
. Thenjz1j=1 .
Let+be the angle for the positivez1and-for the negative. Then+=arctan(-
p
3) =
2
3
and
-=arctan(
p
3) =
4
3
. We can writez1+=e
2i=3
andz1-=e
4i=3
.
z
3
1+
=e
2i
=1
z
3
1-
=e
4i
=1
Therefore,z
3
1
=1 . For the second problem,ij=
3
and
2
3
fori,j= +,-and thejz2j=1 . When we
raisezto the sixth poewr, the argument becomes2and4.
e
2i
=e
4i
=z
6
=1
1.1.2 Square Roots
1.
p
i,
p
-i,
p
1+i,
s
1-i
p
3
2
For
p
i, we are looking forxandysuch that
p
i=x+iy
i=x
2
-y
2
+2xyi
x
2
-y
2
=0 (1.1)
2xy=1 (1.2)
6
u(x,y) =
x
x
2
+y
2
v(x,y) =
-y
x
2
+y
2
For the third problem, we have
x-1+iy
x+1-iy
. Then ¯z=x+1+iy.
x-1+iy
x+1-iy
x+1+iy
x+1+iy
=
x
2
-1+2xyi
(x+1)
2
+y
2
Then real and imaginary parts are
u(x,y) =
x
2
-1
(x+1)
2
+y
2
v(x,y) =
2xy
(x+1)
2
+y
2
For the last problem, we have
1
z
2
=
x
2
-y
2
-2xyi
x
4
+2x
2
y
2
+y
4
so the real and imaginary parts are
u(x,y) =
x
2
-y
2
x
4
+2x
2
y
2
+y
4
v(x,y) =
-2xy
x
4
+2x
2
y
2
+y
4
3.
-1i
p
3
2
3
=1and
1i
p
3
2
6
=1.
Both problems will can be handled easily by converting to polar form. Letz1=
-1i
p
3
2
. Thenjz1j=1 .
Let+be the angle for the positivez1and-for the negative. Then+=arctan(-
p
3) =
2
3
and
-=arctan(
p
3) =
4
3
. We can writez1+=e
2i=3
andz1-=e
4i=3
.
z
3
1+
=e
2i
=1
z
3
1-
=e
4i
=1
Therefore,z
3
1
=1 . For the second problem,ij=
3
and
2
3
fori,j= +,-and thejz2j=1 . When we
raisezto the sixth poewr, the argument becomes2and4.
e
2i
=e
4i
=z
6
=1
1.1.2 Square Roots
1.
p
i,
p
-i,
p
1+i,
s
1-i
p
3
2
For
p
i, we are looking forxandysuch that
p
i=x+iy
i=x
2
-y
2
+2xyi
x
2
-y
2
=0 (1.1)
2xy=1 (1.2)
6
From equation (1.1), we see thatx
2
=y
2 orx=y . Also, note thatiis the upper half plane (UHP).
That is, the angle is positive sox=y and2x
2
=1 from equation (1.1). Therefore,
p
i=
1
p
2
(1+i)
. We
also could have done this problem using the polar form ofz. Letz=i. Thenz=e
i=2 so
p
z=e
i=4
which is exactly what we obtained. For
p
-i
, letz= -i . Thenzin polar form isz=e
-i=2 so
p
z=e
-i=4
=
1
p
2
(1-i)
. For
p
1+i
, letz=1+i . Thenz=
p
2e
i=4
so
p
z=2
1=4
e
i=8
. Finally, for
q
1-i
p
3
2
, letz=
1-i
p
3
2
. Thenz=e
-i=3
so
p
z=e
-i=6
=
1
2
(
p
3-i).
2.
4
p
-1.
Letz=
4
p
-1soz
4
= -1. Letz=re
i
sor
4
e
4i
= -1=e
i(1+2k)
.
r
4
=1
=
4
(1+2k)
wherek=0,1,2,3. Since whenk=4, we havek=0. Then=
4
,
3
4
,
5
4
, and
7
4
.
z=e
i=4
,e
3i=4
,e
5i=4
,e
7i=4
3.
4
p
iand
4
p
-i.
Letz=
4
p
iandz=re
i
. Thenr
4
e
4i
=i=e
i=2
.
r
4
=1
=
8
soz=e
i=8
. Now, letz=
4
p
-i. Thenr
4
e
4i
=e
-i=2
soz=e
-i=8
.
4.
z
2
+ (+i)z++i=0.
The quadratic equation isx=
-b
p
b
2
-ac
2
. For the complex polynomial, we have
z=
--i
p
2
-
2
-4+i(2-4)
2
Leta+bi=
p
2
-
2
-4+i(2-4). Then
z=
--(a+bi)
2
1.1.3 Justication
1.
-
!
,
combined by matrix addition and matrix multiplication, is isomorphic to the eld of complex numbers.
2.Show that the complex number system can be thought of as the eld of all polynomials with real
coefcients modulo the irreducible polynomialx
2
+1.
1.1.4 Conjugation, Absolute Value
1.
z
z
2
+1
forz=x+iyand ¯z=x-iyare conjugate.
7
2
=y
2 orx=y . Also, note thatiis the upper half plane (UHP).
That is, the angle is positive sox=y and2x
2
=1 from equation (1.1). Therefore,
p
i=
1
p
2
(1+i)
. We
also could have done this problem using the polar form ofz. Letz=i. Thenz=e
i=2 so
p
z=e
i=4
which is exactly what we obtained. For
p
-i
, letz= -i . Thenzin polar form isz=e
-i=2 so
p
z=e
-i=4
=
1
p
2
(1-i)
. For
p
1+i
, letz=1+i . Thenz=
p
2e
i=4
so
p
z=2
1=4
e
i=8
. Finally, for
q
1-i
p
3
2
, letz=
1-i
p
3
2
. Thenz=e
-i=3
so
p
z=e
-i=6
=
1
2
(
p
3-i).
2.
4
p
-1.
Letz=
4
p
-1soz
4
= -1. Letz=re
i
sor
4
e
4i
= -1=e
i(1+2k)
.
r
4
=1
=
4
(1+2k)
wherek=0,1,2,3. Since whenk=4, we havek=0. Then=
4
,
3
4
,
5
4
, and
7
4
.
z=e
i=4
,e
3i=4
,e
5i=4
,e
7i=4
3.
4
p
iand
4
p
-i.
Letz=
4
p
iandz=re
i
. Thenr
4
e
4i
=i=e
i=2
.
r
4
=1
=
8
soz=e
i=8
. Now, letz=
4
p
-i. Thenr
4
e
4i
=e
-i=2
soz=e
-i=8
.
4.
z
2
+ (+i)z++i=0.
The quadratic equation isx=
-b
p
b
2
-ac
2
. For the complex polynomial, we have
z=
--i
p
2
-
2
-4+i(2-4)
2
Leta+bi=
p
2
-
2
-4+i(2-4). Then
z=
--(a+bi)
2
1.1.3 Justication
1.
-
!
,
combined by matrix addition and matrix multiplication, is isomorphic to the eld of complex numbers.
2.Show that the complex number system can be thought of as the eld of all polynomials with real
coefcients modulo the irreducible polynomialx
2
+1.
1.1.4 Conjugation, Absolute Value
1.
z
z
2
+1
forz=x+iyand ¯z=x-iyare conjugate.
7
Forz, we have thatz
2
=x
2
-y
2
+2xyi.
z
z
2
+1
=
x+iy
x
2
-y
2
+1+2xyi
=
x+iy
x
2
-y
2
+1+2xyi
x
2
-y
2
+1-2xyi
x
2
-y
2
+1-2xyi
=
x(x
2
-y
2
+1) +2xy
2
+iy(x
2
-y
2
+1-2x
2
)
(x
2
-y
2
+1)
2
+4x
2
y
2
(1.3)
For ¯z, we have that ¯z
2
=x
2
-y
2
-2xyi.
¯z
¯z
2
+1
=
x-iy
x
2
-y
2
+1-2xyi
=
x-iy
x
2
-y
2
+1-2xyi
x
2
-y
2
+1+2xyi
x
2
-y
2
+1+2xyi
=
x(x
2
-y
2
+1) +2xy
2
-iy(x
2
-y
2
+1-2x
2
)
(x
2
-y
2
+1)
2
+4x
2
y
2
(1.4)
Therefore, we have that equations (1.3) and (1.4) are conjugates.
2.
-2i(3+i)(2+4i)(1+i)and
(3+4i)(-1+2i)
(-1-i)(3-i)
.
When we expand the rst problem, we have that
z1= -2i(3+i)(2+4i)(1+i) =32+24i
so
jz1j=
p
32
2
+24
2
=40.
For the second problem, we have that
z2=
(3+4i)(-1+2i)
(-1-i)(3-i)
=2-
3
2
i
so
jz2j=
p
4+9=4=
5
2
.
3.
a-b
1-¯ab
=1
if eitherjaj=1orjbj=1. What exception must be made ifjaj=jbj=1?
Recall thatjzj
2
=z¯z.
1
2
=
a-b
1-¯ab
2
1=
a-b
1-¯ab
a-b
1-¯ab
=
a-b
1-¯ab
¯a-¯b
1-a¯b
=
a¯a-a¯b-¯ab+b¯b
1-¯ab-a¯b+a¯ab¯b
(1.5)
Ifjaj=1, thenjaj
2
=a¯a=1and similarly forjbj
2
=1. Then equation (1.5) becomes
1-a¯b-¯ab+b¯b
1-¯ab-a¯b+b¯b
and
1-a¯b-¯ab+a¯a
1-¯ab-a¯b+a¯a
resepctively which is one. Ifjaj=jbj=1, thenjaj
2
=jbj
2
=1so equation (1.5) can be written as
2-a¯b-¯ab
2-¯ab-a¯b
.
Therefore, we must have thata¯b+¯ab6=2.
8
2
=x
2
-y
2
+2xyi.
z
z
2
+1
=
x+iy
x
2
-y
2
+1+2xyi
=
x+iy
x
2
-y
2
+1+2xyi
x
2
-y
2
+1-2xyi
x
2
-y
2
+1-2xyi
=
x(x
2
-y
2
+1) +2xy
2
+iy(x
2
-y
2
+1-2x
2
)
(x
2
-y
2
+1)
2
+4x
2
y
2
(1.3)
For ¯z, we have that ¯z
2
=x
2
-y
2
-2xyi.
¯z
¯z
2
+1
=
x-iy
x
2
-y
2
+1-2xyi
=
x-iy
x
2
-y
2
+1-2xyi
x
2
-y
2
+1+2xyi
x
2
-y
2
+1+2xyi
=
x(x
2
-y
2
+1) +2xy
2
-iy(x
2
-y
2
+1-2x
2
)
(x
2
-y
2
+1)
2
+4x
2
y
2
(1.4)
Therefore, we have that equations (1.3) and (1.4) are conjugates.
2.
-2i(3+i)(2+4i)(1+i)and
(3+4i)(-1+2i)
(-1-i)(3-i)
.
When we expand the rst problem, we have that
z1= -2i(3+i)(2+4i)(1+i) =32+24i
so
jz1j=
p
32
2
+24
2
=40.
For the second problem, we have that
z2=
(3+4i)(-1+2i)
(-1-i)(3-i)
=2-
3
2
i
so
jz2j=
p
4+9=4=
5
2
.
3.
a-b
1-¯ab
=1
if eitherjaj=1orjbj=1. What exception must be made ifjaj=jbj=1?
Recall thatjzj
2
=z¯z.
1
2
=
a-b
1-¯ab
2
1=
a-b
1-¯ab
a-b
1-¯ab
=
a-b
1-¯ab
¯a-¯b
1-a¯b
=
a¯a-a¯b-¯ab+b¯b
1-¯ab-a¯b+a¯ab¯b
(1.5)
Ifjaj=1, thenjaj
2
=a¯a=1and similarly forjbj
2
=1. Then equation (1.5) becomes
1-a¯b-¯ab+b¯b
1-¯ab-a¯b+b¯b
and
1-a¯b-¯ab+a¯a
1-¯ab-a¯b+a¯a
resepctively which is one. Ifjaj=jbj=1, thenjaj
2
=jbj
2
=1so equation (1.5) can be written as
2-a¯b-¯ab
2-¯ab-a¯b
.
Therefore, we must have thata¯b+¯ab6=2.
8
4.
Find the conditions under which the equationaz+b¯z+c=0 in one complex unknown has exactly one
solution, and compute that solution.
Letz=x+iy. Thenaz+b¯z+c=a(x+iy) +b(x-iy) +c=0.
(a+b)x+c=0 (1.6a)
(a-b)y=0 (1.6b)
Lets consider equation (1.6b). We either have thata=b ory=0. Ifa=b, then WLOG equation (1.6a)
can be written as
x=
-c
2a
andy2R . For xeda,b,c, we have innitely many solutions whena=b sincez=
-c
2a
+iy
fory2R .
Ify=0, then equation (1.6a) can be written as
x=
-c
a+b
.
Therefore,z=xand we have only one solution.
5.
n
X
i=1
aibi
2
=
n
X
i=1
jaij
2
n
X
i=1
jbij
2
-
X
16i6j6n
jai
¯bj-aj
¯bij
2
.
Let's consider
n
X
i=1
aibi
2
+
X
16i6j6n
jai
¯bj-aj
¯bij
2
=
n
X
i=1
jaij
2
n
X
i=1
jbij
2
.
Then we can write the lefthand side as
n
X
i=1
aibi
2
+
X
16i6j6n
jai
¯bj-aj
¯bij
2
=
n
X
i=1
aibi
n
X
j=1
¯aj
¯bj+
X
16i6j6n
(ai
¯bj-aj
¯bi)(¯aibj-¯ajbi)
=
n
X
i,j=1
aibi¯aj
¯bj+
X
i6j
jaij
2
jbjj
2
+jajj
2
jbij
2
-
X
i6j
ai¯ajbi
¯bj+¯aiaj
¯bibj
=
n
X
i=j=1
jaij
2
jbij
2
+
n
X
i6=j
aibi¯aj
¯bj
+
X
i6j
jaij
2
jbjj
2
+jajj
2
jbij
2
-
X
i6j
ai¯ajbi
¯bj+¯aiaj
¯bibj
Fori6=j,
P
n
i6=j
aibi¯aj
¯bj-
P
i<j
ai¯ajbi
¯bj+¯aiaj
¯bibj
=0. Thus, we now have
n
X
i=1
jaij
2
jbij
2
+
X
i6j
jaij
2
jbjj
2
+jajj
2
jbij
2
.
When the indicies of both series on the right hand side coincide,
n
X
i=1
jaij
2
n
X
i=1
jbij
2
=
n
X
i=1
jaij
2
jbij
2
. ( 1.7)
That is, bothaiandbiindex together on the left of side equation (1.7). Whenaiandbidont index
together on the left side of equation (1.7),
n
X
i=1
jaij
2
n
X
i=1
jbij
2
=
X
i6j
jaij
2
jbjj
2
+jajj
2
jbij
2
as was needed to be shown.
9
Find the conditions under which the equationaz+b¯z+c=0 in one complex unknown has exactly one
solution, and compute that solution.
Letz=x+iy. Thenaz+b¯z+c=a(x+iy) +b(x-iy) +c=0.
(a+b)x+c=0 (1.6a)
(a-b)y=0 (1.6b)
Lets consider equation (1.6b). We either have thata=b ory=0. Ifa=b, then WLOG equation (1.6a)
can be written as
x=
-c
2a
andy2R . For xeda,b,c, we have innitely many solutions whena=b sincez=
-c
2a
+iy
fory2R .
Ify=0, then equation (1.6a) can be written as
x=
-c
a+b
.
Therefore,z=xand we have only one solution.
5.
n
X
i=1
aibi
2
=
n
X
i=1
jaij
2
n
X
i=1
jbij
2
-
X
16i6j6n
jai
¯bj-aj
¯bij
2
.
Let's consider
n
X
i=1
aibi
2
+
X
16i6j6n
jai
¯bj-aj
¯bij
2
=
n
X
i=1
jaij
2
n
X
i=1
jbij
2
.
Then we can write the lefthand side as
n
X
i=1
aibi
2
+
X
16i6j6n
jai
¯bj-aj
¯bij
2
=
n
X
i=1
aibi
n
X
j=1
¯aj
¯bj+
X
16i6j6n
(ai
¯bj-aj
¯bi)(¯aibj-¯ajbi)
=
n
X
i,j=1
aibi¯aj
¯bj+
X
i6j
jaij
2
jbjj
2
+jajj
2
jbij
2
-
X
i6j
ai¯ajbi
¯bj+¯aiaj
¯bibj
=
n
X
i=j=1
jaij
2
jbij
2
+
n
X
i6=j
aibi¯aj
¯bj
+
X
i6j
jaij
2
jbjj
2
+jajj
2
jbij
2
-
X
i6j
ai¯ajbi
¯bj+¯aiaj
¯bibj
Fori6=j,
P
n
i6=j
aibi¯aj
¯bj-
P
i<j
ai¯ajbi
¯bj+¯aiaj
¯bibj
=0. Thus, we now have
n
X
i=1
jaij
2
jbij
2
+
X
i6j
jaij
2
jbjj
2
+jajj
2
jbij
2
.
When the indicies of both series on the right hand side coincide,
n
X
i=1
jaij
2
n
X
i=1
jbij
2
=
n
X
i=1
jaij
2
jbij
2
. ( 1.7)
That is, bothaiandbiindex together on the left of side equation (1.7). Whenaiandbidont index
together on the left side of equation (1.7),
n
X
i=1
jaij
2
n
X
i=1
jbij
2
=
X
i6j
jaij
2
jbjj
2
+jajj
2
jbij
2
as was needed to be shown.
9
1.1.5 Inequalities
1.
a-b
1-¯ab
< 1
ifjaj< 1andjbj< 1.
From the properties of the modulus, we have that
a-b
1-¯ab
=
ja-bj
j1-¯abj
=
ja-bj
2
j1-¯abj
2
(1.8)
=
(a-b)(¯a-¯b)
(1-¯ab)(1-a¯b)
=
jaj
2
+jbj
2
-a¯b-¯ab
1+jaj
2
jbj
2
-¯ab-a¯b
<
2-a¯b-¯ab
2-¯ab-a¯b
=1 (1.9)
From equations (1.8) and (1.9), we have
ja-bj
2
j1-¯abj
2
< 1
ja-bj
j1-¯abj
< 1
2.
Cauchy's inequality is
ja1b1+ +anbnj
2
6
ja1j
2
+ +janj
2
jb1j
2
+ +jbnj
2
which can be written more compactly as
n
X
i=1
aibi
2
6
n
X
i=1
jaij
2
n
X
i=1
jbij
2
.
For the base case,i=1, we have
ja1b1j
2
= (a1b1)(¯a1
¯b1) =a1¯a1b1
¯b1=ja1j
2
jb1j
2
so the base case is true. Now let the equality hold for allk-12Z wherek-16n . That is, we assume
that
k-1
X
i=1
aibi
2
6
k-1
X
i=1
jaij
2
k-1
X
i=1
jbij
2
to be true.
k-1
X
i=1
aibi
2
+jakbkj
2
6
k-1
X
i=1
jaij
2
k-1
X
i=1
jbij
2
+jakbkj
2
k
X
i=1
aibi
2
6
k-1
X
i=1
jaij
2
k-1
X
i=1
jbij
2
+ (akbk)(¯ak
¯bk)
=
k-1
X
i=1
jaij
2
k-1
X
i=1
jbij
2
+jakj
2
jbkj
2
=
k
X
i=1
jaij
2
k
X
i=1
jbij
2
Therefore, by the principal of mathematical induction, Cauchy's inequality is true for alln>1 for
n2Z
+
.
10
1.
a-b
1-¯ab
< 1
ifjaj< 1andjbj< 1.
From the properties of the modulus, we have that
a-b
1-¯ab
=
ja-bj
j1-¯abj
=
ja-bj
2
j1-¯abj
2
(1.8)
=
(a-b)(¯a-¯b)
(1-¯ab)(1-a¯b)
=
jaj
2
+jbj
2
-a¯b-¯ab
1+jaj
2
jbj
2
-¯ab-a¯b
<
2-a¯b-¯ab
2-¯ab-a¯b
=1 (1.9)
From equations (1.8) and (1.9), we have
ja-bj
2
j1-¯abj
2
< 1
ja-bj
j1-¯abj
< 1
2.
Cauchy's inequality is
ja1b1+ +anbnj
2
6
ja1j
2
+ +janj
2
jb1j
2
+ +jbnj
2
which can be written more compactly as
n
X
i=1
aibi
2
6
n
X
i=1
jaij
2
n
X
i=1
jbij
2
.
For the base case,i=1, we have
ja1b1j
2
= (a1b1)(¯a1
¯b1) =a1¯a1b1
¯b1=ja1j
2
jb1j
2
so the base case is true. Now let the equality hold for allk-12Z wherek-16n . That is, we assume
that
k-1
X
i=1
aibi
2
6
k-1
X
i=1
jaij
2
k-1
X
i=1
jbij
2
to be true.
k-1
X
i=1
aibi
2
+jakbkj
2
6
k-1
X
i=1
jaij
2
k-1
X
i=1
jbij
2
+jakbkj
2
k
X
i=1
aibi
2
6
k-1
X
i=1
jaij
2
k-1
X
i=1
jbij
2
+ (akbk)(¯ak
¯bk)
=
k-1
X
i=1
jaij
2
k-1
X
i=1
jbij
2
+jakj
2
jbkj
2
=
k
X
i=1
jaij
2
k
X
i=1
jbij
2
Therefore, by the principal of mathematical induction, Cauchy's inequality is true for alln>1 for
n2Z
+
.
10
3.jaij< 1,i>0fori=1,: : :,nand1+2+ +n=1, show that
j1a1+2a2+ +nanj< 1.
Since
P
n
i=1
i=1andi>0,06i< 1. By the triangle inequality,
j1a1+2a2+ +nanj6j1jja1j+ +janjjnj
<
n
X
i=1
i
=1
4. zsatisfying
jz-aj+jz+aj=2jcj
if and only ifjaj6jcj. If this condition is fullled, what are the smallest and largest valuesjzj?
By the triangle inequality,
jz-aj+jz+aj>j(z-a) - (z+a)j=2jaj
so
2jcj=jz-aj+jz+aj
>j(z-a) - (z+a)j
=2jaj
Thus,jcj>jaj. For the second implication, ifa=0, the result follow. Supposea6=0. Then letz=jcj
a
jaj
.
2jcj=jaj(jcj=jaj-1) +jaj(jcj=jaj+1)
=jz-aj+jz+aj
The smallest and largest values ofzcan be found below.
2jcj=jz+aj+jz-aj
4jcj
2
=
jz+aj+jz-aj
2
=2(jzj
2
+jaj
2
)
64(jzj
2
+jaj
2
)
jcj
2
6jzj
2
+jaj
2
q
jcj
2
-jaj
2
6jzj
1.2 The Geometric Representation of Complex Numbers
1.2.1 Geometric Addition and Multiplication
1.
Find the symmetric points ofawith respect to the lines which bisect the angles between the coordinate
axes.
2.
Prove that the pointsa1,a2,a3are vertices of an equilateral triangle if and only ifa
2
1
+a
2
2
+a
2
3
=
a1a2+a2a3+a1a3.
3. aandbare two vertices of a square. Find the two other vertices in all possible cases.
4.
Find the center and the radius of the circle which circumscribes the triangle with verticesa1,a2,a3.
Express the result in symmetric form.
11
j1a1+2a2+ +nanj< 1.
Since
P
n
i=1
i=1andi>0,06i< 1. By the triangle inequality,
j1a1+2a2+ +nanj6j1jja1j+ +janjjnj
<
n
X
i=1
i
=1
4. zsatisfying
jz-aj+jz+aj=2jcj
if and only ifjaj6jcj. If this condition is fullled, what are the smallest and largest valuesjzj?
By the triangle inequality,
jz-aj+jz+aj>j(z-a) - (z+a)j=2jaj
so
2jcj=jz-aj+jz+aj
>j(z-a) - (z+a)j
=2jaj
Thus,jcj>jaj. For the second implication, ifa=0, the result follow. Supposea6=0. Then letz=jcj
a
jaj
.
2jcj=jaj(jcj=jaj-1) +jaj(jcj=jaj+1)
=jz-aj+jz+aj
The smallest and largest values ofzcan be found below.
2jcj=jz+aj+jz-aj
4jcj
2
=
jz+aj+jz-aj
2
=2(jzj
2
+jaj
2
)
64(jzj
2
+jaj
2
)
jcj
2
6jzj
2
+jaj
2
q
jcj
2
-jaj
2
6jzj
1.2 The Geometric Representation of Complex Numbers
1.2.1 Geometric Addition and Multiplication
1.
Find the symmetric points ofawith respect to the lines which bisect the angles between the coordinate
axes.
2.
Prove that the pointsa1,a2,a3are vertices of an equilateral triangle if and only ifa
2
1
+a
2
2
+a
2
3
=
a1a2+a2a3+a1a3.
3. aandbare two vertices of a square. Find the two other vertices in all possible cases.
4.
Find the center and the radius of the circle which circumscribes the triangle with verticesa1,a2,a3.
Express the result in symmetric form.
11
1.2.2 The Binomial Equation
1. (3'), cos(4'), and sin(5')in terms of cos(')and sin(').
For these problems, the sum addition identities will be employed; that is,
cos() =cos()cos()sin()sin()
sin() =sin()cos()sin()cos()
We can write cos(3')as cos(2'+')so
cos(3') =cos(2'+')
=cos(2')cos(') -sin(2')sin(')
=
cos
2
(') -sin
2
(')
cos(') -2sin(')cos(')sin(')
=cos
3
(') -3sin
2
(')cos(')
For cos(4'), we have
cos(4') =cos(2')cos(2') -sin(2')sin(2')
=
cos
2
(') -sin
2
(')
2
-4sin
2
(')cos
2
(')
=cos
4
(') +sin
4
(') -6sin
2
(')cos
2
(')
For sin(5'), we have
sin(5') =sin(4')cos(') +sin(')cos(4')
=2sin(2')cos(2')
cos
2
(') -sin
2
(')
cos(') +sin
5
(') +sin(')cos
4
(') -6sin
3
(')cos
2
(')
=5sin(')cos
4
(') -10sin
3
(')cos
2
(') +sin
5
(')
2. 1+cos(') +cos(2') + +cos(n')and sin(') + +sin(n').
Instead of considering the two separate series, we will consider the series
1+cos(') +isin(') + +cos(n') +isin(n') =1+e
i'
+e
2i'
+ +e
ni'
=
n
X
k=0
e
ki'
Recall that
P
n-1
k=0
r
k
=
1-r
k
1-r
. So
=
1-e
i'(n+1)
1-e
i'
=
e
i'(n+1)
-1
e
i'
-1
(1.10)
Note that sin(
'
2
) =
e
i'=2
-e
-i'=2
2i
so2ie
i'=2
sin(
'
2
) =e
i'
-1. We can now write equation (1.10) as
n
X
k=0
e
ki'
=
e
i'(n+1)=2
sin
'(n+1)
2
e
i'=2
sin
'
2
=
sin
'(n+1)
2
sin
'
2
e
in'=2
(1.11)
By taking the real and imaginary parts of equation (1.11), we get the series for
P
n
k=0
cos(n')
and
P
n
k=0
sin(n'), respectively.
n
X
k=0
cos(n') =
sin
'(n+1)
2
sin
'
2
cos
n'
2
n
X
k=0
sin(n') =
sin
'(n+1)
2
sin
'
2
sin
n'
2
12
1. (3'), cos(4'), and sin(5')in terms of cos(')and sin(').
For these problems, the sum addition identities will be employed; that is,
cos() =cos()cos()sin()sin()
sin() =sin()cos()sin()cos()
We can write cos(3')as cos(2'+')so
cos(3') =cos(2'+')
=cos(2')cos(') -sin(2')sin(')
=
cos
2
(') -sin
2
(')
cos(') -2sin(')cos(')sin(')
=cos
3
(') -3sin
2
(')cos(')
For cos(4'), we have
cos(4') =cos(2')cos(2') -sin(2')sin(2')
=
cos
2
(') -sin
2
(')
2
-4sin
2
(')cos
2
(')
=cos
4
(') +sin
4
(') -6sin
2
(')cos
2
(')
For sin(5'), we have
sin(5') =sin(4')cos(') +sin(')cos(4')
=2sin(2')cos(2')
cos
2
(') -sin
2
(')
cos(') +sin
5
(') +sin(')cos
4
(') -6sin
3
(')cos
2
(')
=5sin(')cos
4
(') -10sin
3
(')cos
2
(') +sin
5
(')
2. 1+cos(') +cos(2') + +cos(n')and sin(') + +sin(n').
Instead of considering the two separate series, we will consider the series
1+cos(') +isin(') + +cos(n') +isin(n') =1+e
i'
+e
2i'
+ +e
ni'
=
n
X
k=0
e
ki'
Recall that
P
n-1
k=0
r
k
=
1-r
k
1-r
. So
=
1-e
i'(n+1)
1-e
i'
=
e
i'(n+1)
-1
e
i'
-1
(1.10)
Note that sin(
'
2
) =
e
i'=2
-e
-i'=2
2i
so2ie
i'=2
sin(
'
2
) =e
i'
-1. We can now write equation (1.10) as
n
X
k=0
e
ki'
=
e
i'(n+1)=2
sin
'(n+1)
2
e
i'=2
sin
'
2
=
sin
'(n+1)
2
sin
'
2
e
in'=2
(1.11)
By taking the real and imaginary parts of equation (1.11), we get the series for
P
n
k=0
cos(n')
and
P
n
k=0
sin(n'), respectively.
n
X
k=0
cos(n') =
sin
'(n+1)
2
sin
'
2
cos
n'
2
n
X
k=0
sin(n') =
sin
'(n+1)
2
sin
'
2
sin
n'
2
12
3.
To nd the roots of unity, we are looking to solvez
n
=1 . Letz=e
i and1=e
2ik . Then=
2k
n
. For
the fth roots of unity,n=5andk=0,1,: : :,4so we have
!0=e
0
=cos(0) +isin(0)
=1
!1=e
2=5
=cos
2
5
+isin
2
5
!2=e
4=5
=cos
4
5
+isin
4
5
!3=e
6=5
=cos
6
5
+isin
6
5
!4=e
8=5
=cos
8
5
+isin
8
5
Now we can plot the roots of unity on the unit circle.
Figure1.1:
For the tenth roots of unity,n=10andk=0,1,: : :,9so we have
!0=e
0
=cos(0) +isin(0)
=1
!1=e
2=10
=cos
5
+isin
5
!2=e
4=10
=cos
2
5
+isin
2
5
!3=e
6=10
=cos
3
5
+isin
3
5
!4=e
8=10
=cos
4
5
+isin
4
5
!5=e
10=10
=cos() +isin()
= -1
!6=e
12=10
=cos
6
5
+isin
6
5
!7=e
14=10
=cos
7
5
+isin
7
5
!8=e
16=10
=cos
8
5
+isin
8
5
!9=e
18=10
=cos
9
5
+isin
9
5
Now we can plot the roots of unity on the unit circle.
13
To nd the roots of unity, we are looking to solvez
n
=1 . Letz=e
i and1=e
2ik . Then=
2k
n
. For
the fth roots of unity,n=5andk=0,1,: : :,4so we have
!0=e
0
=cos(0) +isin(0)
=1
!1=e
2=5
=cos
2
5
+isin
2
5
!2=e
4=5
=cos
4
5
+isin
4
5
!3=e
6=5
=cos
6
5
+isin
6
5
!4=e
8=5
=cos
8
5
+isin
8
5
Now we can plot the roots of unity on the unit circle.
Figure1.1:
For the tenth roots of unity,n=10andk=0,1,: : :,9so we have
!0=e
0
=cos(0) +isin(0)
=1
!1=e
2=10
=cos
5
+isin
5
!2=e
4=10
=cos
2
5
+isin
2
5
!3=e
6=10
=cos
3
5
+isin
3
5
!4=e
8=10
=cos
4
5
+isin
4
5
!5=e
10=10
=cos() +isin()
= -1
!6=e
12=10
=cos
6
5
+isin
6
5
!7=e
14=10
=cos
7
5
+isin
7
5
!8=e
16=10
=cos
8
5
+isin
8
5
!9=e
18=10
=cos
9
5
+isin
9
5
Now we can plot the roots of unity on the unit circle.
13
Figure1.2:
4.!is given by!=cos
2
n
+isin
2
n
, prove that
1+!
h
+!
2h
+ +!
(n-1)h
=0
for any integerhwhich is not a multiple ofn.
Let!=cos
2
n
+isin
2
n
be written in exonential form as!=e
2i=n . Then the series can be written
as
n-1
X
k=0
e
2ih=n
k
=
e
2ih
-1
e
2hi=n
-1
.
Sincehis an integer,e
2ih
=1; therefore, the series zero.
5.
1-!
h
+!
2h
- + (-1)
n-1
!
(n-1)h
?
We can represent this series similarly as
n-1
X
k=0
-e
2ih=n
k
=
(-1)
n
e
2ih
-1
-e
2hi=n
-1
=
1+ (-1)
n+1
e
2ih
1+e
2hi=n
.
Again, sincehis an intger, we have thate
2ih
=1which leaves us with
1+ (-1)
n+1
1+e
2hi=n
=
0, if nis even
2
1+e
2hi=n, ifnis odd
1.2.3 Analytic Geometry
1. az+b¯z+c=0represent a line?
2.
Forx,y,h,k,a,b2Rsuch thata,b6=0, we dene a real ellipse as
(x-h)
2
a
2
+
(y-k)
2
b
2
=1.
Letz=
x
a
+i
y
b
andz0=
h
a
+i
k
b
. If we expand the equation for an ellipse, we have
x
2
a
2
+
y
2
b
2
+
h
2
a
2
+
k
2
b
2
-
2xh
a
2
-
2yk
b
2
=1.
Notice thatjzj
2
=
x
2
a
2+
y
2
b
2andjz0j
2
=
h
2
a
2+
k
2
b
2. Now, let's write the ellipse as
jzj
2
+jz0j
2
-
2xh
a
2
-
2yk
b
2
+
yh
ab
i-
yh
ab
i+
xk
ab
i-
xk
ab
i=jzj
2
+jz0j
2
-¯zz0-z¯z0=1.
Thus, the equation of an ellipse in the complex plane is
(z-z0)(¯z-¯z0) =jz-z0j
2
=1)jz-z0j=1
14
4.!is given by!=cos
2
n
+isin
2
n
, prove that
1+!
h
+!
2h
+ +!
(n-1)h
=0
for any integerhwhich is not a multiple ofn.
Let!=cos
2
n
+isin
2
n
be written in exonential form as!=e
2i=n . Then the series can be written
as
n-1
X
k=0
e
2ih=n
k
=
e
2ih
-1
e
2hi=n
-1
.
Sincehis an integer,e
2ih
=1; therefore, the series zero.
5.
1-!
h
+!
2h
- + (-1)
n-1
!
(n-1)h
?
We can represent this series similarly as
n-1
X
k=0
-e
2ih=n
k
=
(-1)
n
e
2ih
-1
-e
2hi=n
-1
=
1+ (-1)
n+1
e
2ih
1+e
2hi=n
.
Again, sincehis an intger, we have thate
2ih
=1which leaves us with
1+ (-1)
n+1
1+e
2hi=n
=
0, if nis even
2
1+e
2hi=n, ifnis odd
1.2.3 Analytic Geometry
1. az+b¯z+c=0represent a line?
2.
Forx,y,h,k,a,b2Rsuch thata,b6=0, we dene a real ellipse as
(x-h)
2
a
2
+
(y-k)
2
b
2
=1.
Letz=
x
a
+i
y
b
andz0=
h
a
+i
k
b
. If we expand the equation for an ellipse, we have
x
2
a
2
+
y
2
b
2
+
h
2
a
2
+
k
2
b
2
-
2xh
a
2
-
2yk
b
2
=1.
Notice thatjzj
2
=
x
2
a
2+
y
2
b
2andjz0j
2
=
h
2
a
2+
k
2
b
2. Now, let's write the ellipse as
jzj
2
+jz0j
2
-
2xh
a
2
-
2yk
b
2
+
yh
ab
i-
yh
ab
i+
xk
ab
i-
xk
ab
i=jzj
2
+jz0j
2
-¯zz0-z¯z0=1.
Thus, the equation of an ellipse in the complex plane is
(z-z0)(¯z-¯z0) =jz-z0j
2
=1)jz-z0j=1
14
wherezandz0are dened above. Additionally, the standard form of an ellipse in the complex plane is
of the form
jz-aj+jz-bj=c
wherec >ja-bj . Letaandbbe the foci of a hyperbola. Then when the magnitude of the difference of
zand the foci is a constant, we will have a hyperbola.
jz-aj-jz-bj
=c
3.
Prove that the diagonals of a parallelogram bisect each other and that the diagonals of a rhombus are
orthogonal.
4.
Prove analytically that the midpoints of parallel chords to a circle lie on a diameter perpendicular to
the chords.
5. aand1=¯aintersect the circlejzj=1at right angles.
1.2.4 The Spherical Representation
1.
Show thatzandz
0correspond to diametrically opposite points on the Riemann sphere if and only if
z¯z
0
= -1.
2.
A cube has its vertices on the sphereSand its edges parallel to the coordinate axes. Find the stereo-
graphic projections of the vertices.
3.
4.
LetZ,Z
0denote the stereographic projections ofz,z
0, and letNbe the north pole. Show that the
trianglesNZZ
0
andNzz
0
are similar, and use this to derive
d(z,z
0
) =
2jz-z
0
j
p
1+jzj
2
p
1+jz
0
j
2
.
5. aand radiusR.
Letz=a+Randz
0
=a-R. Then the distanced(z,z
0
) =2R.
2R=d(z,z
0
)
R=d(z,z
0
)=2
=
2jRj
p
(1+jaj
2
+jRj
2
+2<fa¯Rg)(1+jaj
2
+jRj
2
-2<fa¯Rg)
=
2jRj
p
(1+jaj
2
+jRj
2
)
2
-4<
2
fa¯Rg
15
of the form
jz-aj+jz-bj=c
wherec >ja-bj . Letaandbbe the foci of a hyperbola. Then when the magnitude of the difference of
zand the foci is a constant, we will have a hyperbola.
jz-aj-jz-bj
=c
3.
Prove that the diagonals of a parallelogram bisect each other and that the diagonals of a rhombus are
orthogonal.
4.
Prove analytically that the midpoints of parallel chords to a circle lie on a diameter perpendicular to
the chords.
5. aand1=¯aintersect the circlejzj=1at right angles.
1.2.4 The Spherical Representation
1.
Show thatzandz
0correspond to diametrically opposite points on the Riemann sphere if and only if
z¯z
0
= -1.
2.
A cube has its vertices on the sphereSand its edges parallel to the coordinate axes. Find the stereo-
graphic projections of the vertices.
3.
4.
LetZ,Z
0denote the stereographic projections ofz,z
0, and letNbe the north pole. Show that the
trianglesNZZ
0
andNzz
0
are similar, and use this to derive
d(z,z
0
) =
2jz-z
0
j
p
1+jzj
2
p
1+jz
0
j
2
.
5. aand radiusR.
Letz=a+Randz
0
=a-R. Then the distanced(z,z
0
) =2R.
2R=d(z,z
0
)
R=d(z,z
0
)=2
=
2jRj
p
(1+jaj
2
+jRj
2
+2<fa¯Rg)(1+jaj
2
+jRj
2
-2<fa¯Rg)
=
2jRj
p
(1+jaj
2
+jRj
2
)
2
-4<
2
fa¯Rg
15
2 Complex Functions
2.1 Introduction to the Concept of Analytical Function
2.1.1 Limits and Continuity
2.1.2 Analytic Functions
1.g(w)andf(z)are analytic functions, show thatg(f(z))is also analytic.
Letg(w) =h(x,y) +it(x,y)andf(z) =u(x,y) +iv(x,y)wherez=w=x+iyforx,y2R. Then
(gf)(z) =h(u(x,y),v(x,y)) +it(u(x,y),v(x,y)).
Sincefandgsatisfy the Cauchy-Riemann equations,
@u
@x
=
@v
@y
@u
@y
= -
@v
@x
@h
@x
=
@t
@y
@h
@u
= -
@t
@x
The partial deivatives of(gf)(z)are
@h
@x
=
@h
@u
@u
@x
+
@t
@v
@v
@x
@t
@y
=
@t
@u
@u
@y
+
@t
@v
@v
@y
@h
@y
=
@h
@u
@u
@y
+
@t
@v
@v
@y
@t
@x
=
@t
@u
@u
@x
+
@t
@v
@v
@x
In order forg(f(z))to be analytic,
@h
@x
=
@t
@y
and
@h
@y
= -
@t
@x
. We can then write
@h
@x
-
@t
@y
=
@h
@u
@u
@x
+
@h
@v
@v
@x
-
@t
@u
@u
@y
-
@t
@v
@v
@y
=
@h
@u
@u
@x
-
@t
@u
@u
@y
| {z }
term1
+
@h
@v
@v
@x
-
@t
@v
@v
@y
| {z }
term2
(2.1)
In order for the right hand side of equation (2.1) to be zero, we need both terms to be zero.
@h
@u
@u
@x
-
@t
@u
@u
@y
=
@h
@u
@u
@x
-
@t
@y
@u
@u
=
@h
@u
@u
@x
-
@h
@u
@u
@x
(2.2)
Equation (2.2) occurs sincegis analytic and satises the Cauchy-Riemann equations.
=0
For the second term in equation (2.1), we again use the analyticity ofg.
@h
@v
@v
@x
-
@t
@v
@v
@y
=
@h
@v
@v
@x
-
@h
@v
@v
@x
=0
Therefore, from equation (2.1), we have
@h
@x
-
@t
@y
=0
@h
@x
=
@t
@y
By similar analysis, we are able to conclude that
@h
@y
= -
@t
@x
. Therefore,g(f(z)) satises the Cauchy-
Riemann so it is analytic.
17
2.1 Introduction to the Concept of Analytical Function
2.1.1 Limits and Continuity
2.1.2 Analytic Functions
1.g(w)andf(z)are analytic functions, show thatg(f(z))is also analytic.
Letg(w) =h(x,y) +it(x,y)andf(z) =u(x,y) +iv(x,y)wherez=w=x+iyforx,y2R. Then
(gf)(z) =h(u(x,y),v(x,y)) +it(u(x,y),v(x,y)).
Sincefandgsatisfy the Cauchy-Riemann equations,
@u
@x
=
@v
@y
@u
@y
= -
@v
@x
@h
@x
=
@t
@y
@h
@u
= -
@t
@x
The partial deivatives of(gf)(z)are
@h
@x
=
@h
@u
@u
@x
+
@t
@v
@v
@x
@t
@y
=
@t
@u
@u
@y
+
@t
@v
@v
@y
@h
@y
=
@h
@u
@u
@y
+
@t
@v
@v
@y
@t
@x
=
@t
@u
@u
@x
+
@t
@v
@v
@x
In order forg(f(z))to be analytic,
@h
@x
=
@t
@y
and
@h
@y
= -
@t
@x
. We can then write
@h
@x
-
@t
@y
=
@h
@u
@u
@x
+
@h
@v
@v
@x
-
@t
@u
@u
@y
-
@t
@v
@v
@y
=
@h
@u
@u
@x
-
@t
@u
@u
@y
| {z }
term1
+
@h
@v
@v
@x
-
@t
@v
@v
@y
| {z }
term2
(2.1)
In order for the right hand side of equation (2.1) to be zero, we need both terms to be zero.
@h
@u
@u
@x
-
@t
@u
@u
@y
=
@h
@u
@u
@x
-
@t
@y
@u
@u
=
@h
@u
@u
@x
-
@h
@u
@u
@x
(2.2)
Equation (2.2) occurs sincegis analytic and satises the Cauchy-Riemann equations.
=0
For the second term in equation (2.1), we again use the analyticity ofg.
@h
@v
@v
@x
-
@t
@v
@v
@y
=
@h
@v
@v
@x
-
@h
@v
@v
@x
=0
Therefore, from equation (2.1), we have
@h
@x
-
@t
@y
=0
@h
@x
=
@t
@y
By similar analysis, we are able to conclude that
@h
@y
= -
@t
@x
. Therefore,g(f(z)) satises the Cauchy-
Riemann so it is analytic.
17
2. z
2
andz
3
.
Letz=x+iy . Thenz
2
=x
2
-y
2
+2xyi andz
3
=x
3
-3xy
2
+i(3x
2
y-y
3
) . Forf(z) =z
2 , the Cauchy-
Riemann equations are
ux=2x v y=2x
uy= -2y -vx= -2y
Thus, the Cauchy-Riemann equation satised forf(z) =z
2 . Forf(z) =z
3 , the Cauchy-Riemann
equations are
ux=3x
2
-3y
2
vy=3x
2
-3y
2
uy= -6xy -vx= -6xy
Thus, the Cauchy-Riemann equation satised forf(z) =z
3
.
3.
Find the most general harmonic polynomial of the formax
3
+bx
2
y+cxy
2
+dy
3 . Determine the
conjugate harmonic function and the corresponding analytic function by integration and by the formal
method.
In order to be harmonic,u(x,y) =ax
3
+bx
2
y+cxy
2
+dy
3
has to satisfyr
2
u=0so
uxx+uyy= (3a+c)x+ (3d+b)y=0.
Thus,3a= -cand3d= -bso
u(x,y) =ax
3
-3axy
2
-3dx
2
y+dy
3
.
To nd the harmonic conjugatev(x,y, we need to look at the Cauchy-Riemann equations. By the
Cauchy-Riemann equations,
ux=3ax
2
-3ay
2
-6dxy=vy.
Then we can integrate with respect toyto ndv(x,y).
v(x,y) =
Z
(3ax
2
-3ay
2
-6dxy)dy=3ax
2
y-ay
3
-3dxy
2
+g(x)
Using the second Cauchy-Riemann, we have
vx=6axy-3dy
2
+g
0
(x) = -uy=3dx
2
+6axy-3dy
2
sog
0
(x) =3dx
2
. Theng(x) =dx
3
+Cand
v(x,y) =3ax
2
y-ay
3
-3dxy
2
+dx
3
+C.
4.
Letf=u(x ,y) +iv(x ,y). Then the modulus offisjfj=
p
u
2
+v
2
. If the modulus offis constant, then
u
2
+v
2
=c for some constantc. Ifc=0, thenf=0 which is constant. Supposec6=0. By taking the
derivative with respect toxandy, we have
0=
@
@x
(u
2
+v
2
)
=2uux+2vvx
=uux+vvx
0=
@
@y
(u
2
+v
2
)
=uuy+vvy
Sincefis analytic,fsatises the Cauchy-Riemann. That is,ux=vyanduy= -vx.
uux-vuy=0 (2.3a)
18
2
andz
3
.
Letz=x+iy . Thenz
2
=x
2
-y
2
+2xyi andz
3
=x
3
-3xy
2
+i(3x
2
y-y
3
) . Forf(z) =z
2 , the Cauchy-
Riemann equations are
ux=2x v y=2x
uy= -2y -vx= -2y
Thus, the Cauchy-Riemann equation satised forf(z) =z
2 . Forf(z) =z
3 , the Cauchy-Riemann
equations are
ux=3x
2
-3y
2
vy=3x
2
-3y
2
uy= -6xy -vx= -6xy
Thus, the Cauchy-Riemann equation satised forf(z) =z
3
.
3.
Find the most general harmonic polynomial of the formax
3
+bx
2
y+cxy
2
+dy
3 . Determine the
conjugate harmonic function and the corresponding analytic function by integration and by the formal
method.
In order to be harmonic,u(x,y) =ax
3
+bx
2
y+cxy
2
+dy
3
has to satisfyr
2
u=0so
uxx+uyy= (3a+c)x+ (3d+b)y=0.
Thus,3a= -cand3d= -bso
u(x,y) =ax
3
-3axy
2
-3dx
2
y+dy
3
.
To nd the harmonic conjugatev(x,y, we need to look at the Cauchy-Riemann equations. By the
Cauchy-Riemann equations,
ux=3ax
2
-3ay
2
-6dxy=vy.
Then we can integrate with respect toyto ndv(x,y).
v(x,y) =
Z
(3ax
2
-3ay
2
-6dxy)dy=3ax
2
y-ay
3
-3dxy
2
+g(x)
Using the second Cauchy-Riemann, we have
vx=6axy-3dy
2
+g
0
(x) = -uy=3dx
2
+6axy-3dy
2
sog
0
(x) =3dx
2
. Theng(x) =dx
3
+Cand
v(x,y) =3ax
2
y-ay
3
-3dxy
2
+dx
3
+C.
4.
Letf=u(x ,y) +iv(x ,y). Then the modulus offisjfj=
p
u
2
+v
2
. If the modulus offis constant, then
u
2
+v
2
=c for some constantc. Ifc=0, thenf=0 which is constant. Supposec6=0. By taking the
derivative with respect toxandy, we have
0=
@
@x
(u
2
+v
2
)
=2uux+2vvx
=uux+vvx
0=
@
@y
(u
2
+v
2
)
=uuy+vvy
Sincefis analytic,fsatises the Cauchy-Riemann. That is,ux=vyanduy= -vx.
uux-vuy=0 (2.3a)
18
uuy+vux=0 (2.3b)
Let's write equations (2.3a) and (2.3b) in matrix form. Then we have
"
u-v
v u
# "
@u
@x
@u
@x
#
=
"
0
0
#
Suppose the matrix is not invertible. Thenu
2
+v
2
=0 . Sinceu
2,v
2
2R ,u
2,v
2
>0 . Therefore,u=v=0
sof(z) =0. Now, suppose that the matrix is invertible. Then we have
"
@u
@x
@u
@x
#
=
"
0
0
#
sof
0
(z) =0andf(z) =cfor some constantc.
5. f(z)andf(¯z)are simultaneously analytic.
Letg(z) =f(¯z)and supposefis analytic. Theng
0
(z)is
g
0
(z) =lim
z!0
g(z+z) -g(z)
z
=lim
z!0
f(¯z+z) -f(¯z)
z
=lim
z!0
f(¯z+z) -f(¯z)
z
Since conjugation is continuous, we can move the limit inside the conjugation.
=lim
z!0
f(¯z+z) -f(¯z)
z
=f
0
(¯z)
Thus,gis differentiable with derivative
f
0
(¯z)
. Suppose
f(¯z)
is analytic and let
g(¯z) =f(z)
. Then by the
same argument,fis differentiable with derivative
g
0
(¯z)
. Therefore,f(z)and
f(¯z)
are simultaneously
analytic.
We could also use the Cauchy-Riemann equations. Letf(z) =u(x ,y) +iv(x ,y)wherez=x+iy so
¯z=x-iy . Then
f(¯z) =(x
,y) -i(x ,y)where(x,y) =u(x ,-y)and(x,y) =v(x ,-y). In order for
both to be analytic, they both need to satisfy the Cauchy-Riemann equations. That is,ux=vy ,uy= -vx ,
x=yandy= -x.
ux(x,y) =vy(x,y)
uy(x,y) = -vx(x,y)
x(x,y) =ux(x,-y)
y(x,y) = -uy(x,-y)
-x(x,y) =vx(x,-y)
y(x,y) =vy(x,-y)
Suppose that
f(¯z)
satises the Cauchy-Riemann equations. Thenx=ux(x ,-y) =vy(x ,-y) =y and
y= -uy(x,-y) =vx(x,-y) = -x. Therefore,
ux(x,-y) =vy(x,-y)
uy(x,-y) = -vx(x,-y)
which meansf(¯z)satises the Cauchy-Riemann equations. Now, recall thatjzj=j¯zj . Sincef(¯z)satises
the Cauchy-Riemann equations, for an > 0there exists a > 0such that when0 <jzj< ,
jf(¯z) -¯z0j=jf(z) -z0j< . Thus, limz!0f(z) =z0sof(z)is analytic if
f(¯z)is analytic.
19
Let's write equations (2.3a) and (2.3b) in matrix form. Then we have
"
u-v
v u
# "
@u
@x
@u
@x
#
=
"
0
0
#
Suppose the matrix is not invertible. Thenu
2
+v
2
=0 . Sinceu
2,v
2
2R ,u
2,v
2
>0 . Therefore,u=v=0
sof(z) =0. Now, suppose that the matrix is invertible. Then we have
"
@u
@x
@u
@x
#
=
"
0
0
#
sof
0
(z) =0andf(z) =cfor some constantc.
5. f(z)andf(¯z)are simultaneously analytic.
Letg(z) =f(¯z)and supposefis analytic. Theng
0
(z)is
g
0
(z) =lim
z!0
g(z+z) -g(z)
z
=lim
z!0
f(¯z+z) -f(¯z)
z
=lim
z!0
f(¯z+z) -f(¯z)
z
Since conjugation is continuous, we can move the limit inside the conjugation.
=lim
z!0
f(¯z+z) -f(¯z)
z
=f
0
(¯z)
Thus,gis differentiable with derivative
f
0
(¯z)
. Suppose
f(¯z)
is analytic and let
g(¯z) =f(z)
. Then by the
same argument,fis differentiable with derivative
g
0
(¯z)
. Therefore,f(z)and
f(¯z)
are simultaneously
analytic.
We could also use the Cauchy-Riemann equations. Letf(z) =u(x ,y) +iv(x ,y)wherez=x+iy so
¯z=x-iy . Then
f(¯z) =(x
,y) -i(x ,y)where(x,y) =u(x ,-y)and(x,y) =v(x ,-y). In order for
both to be analytic, they both need to satisfy the Cauchy-Riemann equations. That is,ux=vy ,uy= -vx ,
x=yandy= -x.
ux(x,y) =vy(x,y)
uy(x,y) = -vx(x,y)
x(x,y) =ux(x,-y)
y(x,y) = -uy(x,-y)
-x(x,y) =vx(x,-y)
y(x,y) =vy(x,-y)
Suppose that
f(¯z)
satises the Cauchy-Riemann equations. Thenx=ux(x ,-y) =vy(x ,-y) =y and
y= -uy(x,-y) =vx(x,-y) = -x. Therefore,
ux(x,-y) =vy(x,-y)
uy(x,-y) = -vx(x,-y)
which meansf(¯z)satises the Cauchy-Riemann equations. Now, recall thatjzj=j¯zj . Sincef(¯z)satises
the Cauchy-Riemann equations, for an > 0there exists a > 0such that when0 <jzj< ,
jf(¯z) -¯z0j=jf(z) -z0j< . Thus, limz!0f(z) =z0sof(z)is analytic if
f(¯z)is analytic.
19
6. u(z)andu(¯z)are simultaneously harmonic.
Sinceuis the real part off(z),u(z) =u(x ,y)wherez=x+iy . Supposeu(z)is harmonic. Thenu(z)
satises Laplace equation.
r
2
u(z) =uxx+uyy=0
Now,u(¯z) =u(x,-y)where
@
2
@x
2u(¯z) =uxxand
@
2
@y
2u(¯z) =uyyso
r
2
u(¯z) =uxx+uyy=0.
Sinceu(z)is harmonic,uxx+uyy=0so it follows thatu(¯z)is harmonic as well.
7.
@
2
u
@z@¯z
=0.
Letube a harmonic. Thenr
2
u=0.
@
@¯z
=
1
2
@
@x
+i
@
@y
(2.4a)
@
@z
=
1
2
@
@x
-i
@
@y
(2.4b)
From equation (2.4a), we have
1
2
@
@x
+i
@
@y
u=
1
2
(ux+iuy).
Then we have
@
2
u
@z@¯z
=
1
4
@
@x
-i
@
@y
(ux+iuy) =
1
4
uxx+uyy+i(uyx-uxy)
Sinceuis a solution to the Laplace equation,uhas continuous rst and second derivatives. That is,
u2C
2
at a minimum. By Schwarz's theorem,uxy=uyxso
@
2
u
@z@¯z
=0.
Schwarz's theorem states that iffis a function of two variables such thatfxyandfyxboth exist and are
continuous at some point(x0,y0), thenfxy=fyx.
2.1.3 Polynomials
2.1.4 Rational Functions
1.
z
4
z
3
-1
and
1
z(z+1)
2
(z+2)
3
in partial fractions.
LetR(z) =
z
4
z
3
-1
=z+
z
z
3
-1
. The poles ofR(z)occur whenz
3
=1 . Then the distinct poles are
z=1,e
2i=3
,e
4i=3
. LetH(z) =
z
z
3
-1
,z7!i+1=w, andi2f1,e
2i=3
,e
4i=3
g.
H(1+1=w) =
w
3
-
w
3(3w
2
+3w+1)
H(e
2i=3
+1=w) =
w
3e
2i=3
-
w
3e
2i=3
(3e
2i=3
w
2
+3e
4i=3
w+1)
H(e
4i=3
+1=w) =
w
3e
4i=3
-
w
3e
4i=3
(3e
4i=3
w
2
+3e
2i=3
w+1)
H(i+1=w) =
w
3i
-Q(w)
20
Sinceuis the real part off(z),u(z) =u(x ,y)wherez=x+iy . Supposeu(z)is harmonic. Thenu(z)
satises Laplace equation.
r
2
u(z) =uxx+uyy=0
Now,u(¯z) =u(x,-y)where
@
2
@x
2u(¯z) =uxxand
@
2
@y
2u(¯z) =uyyso
r
2
u(¯z) =uxx+uyy=0.
Sinceu(z)is harmonic,uxx+uyy=0so it follows thatu(¯z)is harmonic as well.
7.
@
2
u
@z@¯z
=0.
Letube a harmonic. Thenr
2
u=0.
@
@¯z
=
1
2
@
@x
+i
@
@y
(2.4a)
@
@z
=
1
2
@
@x
-i
@
@y
(2.4b)
From equation (2.4a), we have
1
2
@
@x
+i
@
@y
u=
1
2
(ux+iuy).
Then we have
@
2
u
@z@¯z
=
1
4
@
@x
-i
@
@y
(ux+iuy) =
1
4
uxx+uyy+i(uyx-uxy)
Sinceuis a solution to the Laplace equation,uhas continuous rst and second derivatives. That is,
u2C
2
at a minimum. By Schwarz's theorem,uxy=uyxso
@
2
u
@z@¯z
=0.
Schwarz's theorem states that iffis a function of two variables such thatfxyandfyxboth exist and are
continuous at some point(x0,y0), thenfxy=fyx.
2.1.3 Polynomials
2.1.4 Rational Functions
1.
z
4
z
3
-1
and
1
z(z+1)
2
(z+2)
3
in partial fractions.
LetR(z) =
z
4
z
3
-1
=z+
z
z
3
-1
. The poles ofR(z)occur whenz
3
=1 . Then the distinct poles are
z=1,e
2i=3
,e
4i=3
. LetH(z) =
z
z
3
-1
,z7!i+1=w, andi2f1,e
2i=3
,e
4i=3
g.
H(1+1=w) =
w
3
-
w
3(3w
2
+3w+1)
H(e
2i=3
+1=w) =
w
3e
2i=3
-
w
3e
2i=3
(3e
2i=3
w
2
+3e
4i=3
w+1)
H(e
4i=3
+1=w) =
w
3e
4i=3
-
w
3e
4i=3
(3e
4i=3
w
2
+3e
2i=3
w+1)
H(i+1=w) =
w
3i
-Q(w)
20
ThenGi(w) =
w
3i
wherew7!1=(z-i). Now,R(z) =G(z) +Gi[1=(z-i)]so we nally have that
R(z) =
z
4
z
3
-1
=z+
z
z
3
-1
=z+
3
X
i=1
1
3i(z-i)
.
The second problem's numerator already is of a degree less than the denominator so we can proceed
at once. LetR(z) =
1
z(z+1)
2
(z+2)
3. The poles ofR(z)arei2f0,-1,-2gandz7!i+1=w.
R(1=w) =
w
(1=w+1)
2
(1=w+2)
3
=
w
6
(w+1)
2
(2w+1)
3
=
w
8
+Q(w)
R(1=w-1) =
w
6
(1-w)(w+1)
3
=2w-w
2
+Q(w)
R(1=w-2) =
w
6
(1-2w)(w-1)
2
= -
17w
8
-
5w
2
4
-
w
3
2
+Q(w)
Therefore, we can write
R(z) =
1
z(z+1)
2
(z+2)
3
=
1
8z
+
2
z+1
-
1
(z+1)
2
-
17
8(z+2)
-
5
4(z+2)
2
-
1
2(z+2)
3
2.
Use the formula in the preceding exercise to prove that there exists a unique polynomialPof degree
< nwith given valuesckat the pointsk(Lagrange's interpolation polynomial).
3.
What is the general form of a rational function which has absolute value1on the circlejzj=1 ? In
particular, how are the zeros and poles related to each other?
4. jzj=1, how are the zeros and poles situated?
5.R(z)is a rational function of ordern, how large and how small can the order ofR
0
(z)be?
LetR(z) =P(z)=Q(z) whereP(z)has degreenandQ(z)has degreem. Letkbe the degree ofR(z). Then
k=maxfn,mg.
R
0
(z) =
P
0
(z)Q(z) -P(z)Q
0
(z)
Q(z)
2
Then we have four cases
(a) P(z)andQ(z)are nonconstant.
ThenP
0
(z)andQ
0
(z) have degreesn-1 andm-1, respectively. Since we are looking only for the
highest degree terms, we have
R
0
(z) =
z
n-1
z
m
-z
n
z
m-1
z
2m
Therefore, the degree ofR
0
(z)isk
0
=maxfn+m-1,2mg.
(b) P(z)is nonconstant andQ(z)is a nonzero constant function.
R
0
(z) =
z
n-1
-z
n
0
z
02
The degree ofR
0
(z)isk
0
=n-1.
(c) P(z)is a nonzero constant function andQ(z)is a nonconstant.
R
0
(z) =
0z
m
-z
m-1
z
2m
The degree ofR
0
(z)isk
0
=2m.
(d) P(z)andQ(z)are nonzero constant functions.
In this case,R
0
(z) =0soP(z) =aandQ(z) =b. Then the degree ofR
0
(z)isk
0
=0.
21
w
3i
wherew7!1=(z-i). Now,R(z) =G(z) +Gi[1=(z-i)]so we nally have that
R(z) =
z
4
z
3
-1
=z+
z
z
3
-1
=z+
3
X
i=1
1
3i(z-i)
.
The second problem's numerator already is of a degree less than the denominator so we can proceed
at once. LetR(z) =
1
z(z+1)
2
(z+2)
3. The poles ofR(z)arei2f0,-1,-2gandz7!i+1=w.
R(1=w) =
w
(1=w+1)
2
(1=w+2)
3
=
w
6
(w+1)
2
(2w+1)
3
=
w
8
+Q(w)
R(1=w-1) =
w
6
(1-w)(w+1)
3
=2w-w
2
+Q(w)
R(1=w-2) =
w
6
(1-2w)(w-1)
2
= -
17w
8
-
5w
2
4
-
w
3
2
+Q(w)
Therefore, we can write
R(z) =
1
z(z+1)
2
(z+2)
3
=
1
8z
+
2
z+1
-
1
(z+1)
2
-
17
8(z+2)
-
5
4(z+2)
2
-
1
2(z+2)
3
2.
Use the formula in the preceding exercise to prove that there exists a unique polynomialPof degree
< nwith given valuesckat the pointsk(Lagrange's interpolation polynomial).
3.
What is the general form of a rational function which has absolute value1on the circlejzj=1 ? In
particular, how are the zeros and poles related to each other?
4. jzj=1, how are the zeros and poles situated?
5.R(z)is a rational function of ordern, how large and how small can the order ofR
0
(z)be?
LetR(z) =P(z)=Q(z) whereP(z)has degreenandQ(z)has degreem. Letkbe the degree ofR(z). Then
k=maxfn,mg.
R
0
(z) =
P
0
(z)Q(z) -P(z)Q
0
(z)
Q(z)
2
Then we have four cases
(a) P(z)andQ(z)are nonconstant.
ThenP
0
(z)andQ
0
(z) have degreesn-1 andm-1, respectively. Since we are looking only for the
highest degree terms, we have
R
0
(z) =
z
n-1
z
m
-z
n
z
m-1
z
2m
Therefore, the degree ofR
0
(z)isk
0
=maxfn+m-1,2mg.
(b) P(z)is nonconstant andQ(z)is a nonzero constant function.
R
0
(z) =
z
n-1
-z
n
0
z
02
The degree ofR
0
(z)isk
0
=n-1.
(c) P(z)is a nonzero constant function andQ(z)is a nonconstant.
R
0
(z) =
0z
m
-z
m-1
z
2m
The degree ofR
0
(z)isk
0
=2m.
(d) P(z)andQ(z)are nonzero constant functions.
In this case,R
0
(z) =0soP(z) =aandQ(z) =b. Then the degree ofR
0
(z)isk
0
=0.
21
2.2 Elementary Theory of Power Series
2.2.1 Sequences
2.2.2 Series
2.2.3 Uniform Convergence
1.
Letfangbe a convergent sequence andlimn!1an=a . Let=1. Then there exists ann > Nsuch that
jan-aj< 1.
janj=jan-a+aj
By the triangle inequality, we have
6jan-aj+jaj
janj-jaj6jan-aj
Therefore, we have that
janj-jaj6jan-aj< 1
janj< 1+jaj
For alln > N,janj< 1+jaj so letA=max
1+jaj ,ja1j,: : :,jaNj
. Thus,janj< A for some niteAand
hencefangis bounded byA.
2. n!1zn=A, prove that
lim
n!1
1
n
(z1+z2+ +zn) =A.
Given > 0there exists somen > Nsuch that
jzn-Aj<
N
2
.
Now, since limn!1zn=Aconverges, it is Cauchy. Therefore, there existsn,m > Nsuch that
jzm-znj<
N
2
.
Repeating this we have thatjz1+ +zn-nAj orj1=n(z1+ +zn-1) -A+ (zn-A)=nj . For a xedN,
we can ndnsuch that
n-1
X
i=1
jzi-Aj<
N
2
We now have that
j1=n(z1+ +zn-1) -A+ (zn-A)=nj6
1=n
n-1
X
i=1
(zi-A)
+1=njzn-Aj (2.5)
61=n
n-1
X
i=1
jzi-Aj+1=njzn-Aj
< 1=n
N
2
+1=n
N
2
<
Equation (2.5) can be written asj1=n(z1+ +zn) -Aj< so
lim
n!1
1=n(z1+ +zn) =A.
22
2.2.1 Sequences
2.2.2 Series
2.2.3 Uniform Convergence
1.
Letfangbe a convergent sequence andlimn!1an=a . Let=1. Then there exists ann > Nsuch that
jan-aj< 1.
janj=jan-a+aj
By the triangle inequality, we have
6jan-aj+jaj
janj-jaj6jan-aj
Therefore, we have that
janj-jaj6jan-aj< 1
janj< 1+jaj
For alln > N,janj< 1+jaj so letA=max
1+jaj ,ja1j,: : :,jaNj
. Thus,janj< A for some niteAand
hencefangis bounded byA.
2. n!1zn=A, prove that
lim
n!1
1
n
(z1+z2+ +zn) =A.
Given > 0there exists somen > Nsuch that
jzn-Aj<
N
2
.
Now, since limn!1zn=Aconverges, it is Cauchy. Therefore, there existsn,m > Nsuch that
jzm-znj<
N
2
.
Repeating this we have thatjz1+ +zn-nAj orj1=n(z1+ +zn-1) -A+ (zn-A)=nj . For a xedN,
we can ndnsuch that
n-1
X
i=1
jzi-Aj<
N
2
We now have that
j1=n(z1+ +zn-1) -A+ (zn-A)=nj6
1=n
n-1
X
i=1
(zi-A)
+1=njzn-Aj (2.5)
61=n
n-1
X
i=1
jzi-Aj+1=njzn-Aj
< 1=n
N
2
+1=n
N
2
<
Equation (2.5) can be written asj1=n(z1+ +zn) -Aj< so
lim
n!1
1=n(z1+ +zn) =A.
22
3.
Let
P
an
be an absolutely convergent series and
P
bn
be its rearrangement. Since
P
an
converges
absolutely, for > 0, there exists an > Nsuch thatjsn-Aj< =2 wheresnis thenth partial sum. Let
tnbe thenth partial sum of
P
bn. Then for somen > N
jtn-Aj=jtn-sn+sn-Aj
6jtn-snj+jsn-Aj
<jtn-snj+
2
Since
P
an
is absolutely convergent,
P
1
k=n+1
jakj
converges to zero. Let the remainder bern. Then for
someN > n,n1,jrn-0j< =2. LetM=maxfk1,k2,: : :,kNg. Then for somen > M, we have
jtn-snj=
N
X
n
an
6
X
janj6
1
X
k=n+1
janj=rn<
2
Thus,jtn-snj< and a rearrangement of an absolutely convergent series does not changes its sum.
4. fnz
n
g
1
n=1
.
Consider whenjzj< 1. Thenz
n
=
1
w
nwherejwj> 1. By the ratio test, we have
lim
n!1
(n+1)w
n
nw
n+1
=
1
jwj
lim
n!1
n+1
n
=
1
jwj
In order for convergence, the ratio test has to be less than one.
1
jwj
=jzj
which is less than one by our assumption sofnz
n
g converges absolutely in the disc less than one. Now,
let's considerjzj>1 . By the ratio test, we getlimn!1jan+1=anj=jzj>1 by our assumption. When the
limit is one, we can draw no conclusion about convergence, but when the limit is greater than one, the
sequence diverges. Forjzj< 1, > 0, andn > N,jnz
n
-0j< for uniform convergence. Takez=9=10,
n=100, and=0.001. Then
jnz
n
j=njzj
n
<
jzj
n
<
n
0.0000266< 0.00001
Thus, the sequence is not uniformly convergent in the disc with radius less than one. Let's consider the
closed discjzj6R whereR2(0 ,1). Nowjnz
n
jis bounded above by a convergent geometric series, say
P
r
n
wherejrj< 1. Thenjnz
n
j< ar
n forjzj6R andaa real constant. LetMn=ar
n whereMnis the
Min the Weierstrass M-test. Thus,fnz
n
gis uniformly convergent in a closed disc less than one.
5.
1
X
n=1
x
n(1+nx
2
)
for real values ofx.
By the AM-GM inequality,(x+y)=2>
p
xy, we have
1+nx
2
>2jxj
p
n
or
1
2jxj
p
n
>
1
1+nx
2. Letfn(x) =
x
n(1+nx
2
)
. Then
jfn(x)j6
x
2xn
3=2
=
1
2n
3=2
=Mn
23
Let
P
an
be an absolutely convergent series and
P
bn
be its rearrangement. Since
P
an
converges
absolutely, for > 0, there exists an > Nsuch thatjsn-Aj< =2 wheresnis thenth partial sum. Let
tnbe thenth partial sum of
P
bn. Then for somen > N
jtn-Aj=jtn-sn+sn-Aj
6jtn-snj+jsn-Aj
<jtn-snj+
2
Since
P
an
is absolutely convergent,
P
1
k=n+1
jakj
converges to zero. Let the remainder bern. Then for
someN > n,n1,jrn-0j< =2. LetM=maxfk1,k2,: : :,kNg. Then for somen > M, we have
jtn-snj=
N
X
n
an
6
X
janj6
1
X
k=n+1
janj=rn<
2
Thus,jtn-snj< and a rearrangement of an absolutely convergent series does not changes its sum.
4. fnz
n
g
1
n=1
.
Consider whenjzj< 1. Thenz
n
=
1
w
nwherejwj> 1. By the ratio test, we have
lim
n!1
(n+1)w
n
nw
n+1
=
1
jwj
lim
n!1
n+1
n
=
1
jwj
In order for convergence, the ratio test has to be less than one.
1
jwj
=jzj
which is less than one by our assumption sofnz
n
g converges absolutely in the disc less than one. Now,
let's considerjzj>1 . By the ratio test, we getlimn!1jan+1=anj=jzj>1 by our assumption. When the
limit is one, we can draw no conclusion about convergence, but when the limit is greater than one, the
sequence diverges. Forjzj< 1, > 0, andn > N,jnz
n
-0j< for uniform convergence. Takez=9=10,
n=100, and=0.001. Then
jnz
n
j=njzj
n
<
jzj
n
<
n
0.0000266< 0.00001
Thus, the sequence is not uniformly convergent in the disc with radius less than one. Let's consider the
closed discjzj6R whereR2(0 ,1). Nowjnz
n
jis bounded above by a convergent geometric series, say
P
r
n
wherejrj< 1. Thenjnz
n
j< ar
n forjzj6R andaa real constant. LetMn=ar
n whereMnis the
Min the Weierstrass M-test. Thus,fnz
n
gis uniformly convergent in a closed disc less than one.
5.
1
X
n=1
x
n(1+nx
2
)
for real values ofx.
By the AM-GM inequality,(x+y)=2>
p
xy, we have
1+nx
2
>2jxj
p
n
or
1
2jxj
p
n
>
1
1+nx
2. Letfn(x) =
x
n(1+nx
2
)
. Then
jfn(x)j6
x
2xn
3=2
=
1
2n
3=2
=Mn
23
For a xedx,
P
jfn(x)j6Mn<1
so
P
jfn(x)j
is absolutely convergent. Thus,
P
fn(x)
is pointwise
convergent tof(x). Let > 0be given andsn=
P
n
k=1
fk(x)benth partial sum. Letn > Nsuch that
jf(x) -snj=
1
X
k=1
fk(x) -
n
X
k=1
fk(x)
=
1
X
k=n+1
fk(x)
6
1
X
k=n+1
jfk(x)j
Since
P
Mk
converges to some limit, fornsufciently large,
P
1
k=n+1
Mk<
. SelectNsuch that this is
true. Then
jf(x) -snj6
1
X
k=n+1
jfk(x)j6
1
X
k=n+1
Mk<
Therefore,
P
fn(x)wherefn(x) =
x
n(1+nx
2
)
is uniformly convergent by the Weierstrass M-test.
6.
IfU=u1+u2+ ,V=v1+v2+ are convergent series, prove thatUV=u1v1+ (u1v2+u2v2) +
(u1v3+u2v2+u3v1) +
provided that at least one of the series is absolutely convergent. (It is easy
if both series are absolutely convergent. Try to rearrange the proof so economically that the absolute
convergence of the second series is not needed.)
2.2.4 Power Series
1. (1-z)
-m
,ma positive integer, in powers ofz.
The Binomial theorem states that(1+x)
n
=
P
1
k=0
n
k
x
k
. In our case, we have
(1-z)
-m
=
1
X
k=0
-m
k
(-z)
k
(2.6)
where
-m
k
= (-1)
k
m+k-1
k
. Then equation (2.6) can be written as
(1-z)
-m
=
1
X
k=0
m+k-1
k
z
k
=1+mz+
m(m+1)
2!
z
2
+ .
2.
2z+3
z+1
in powers ofz-1. What is the radius of convergence?
Let's just consider
1
z+1
for the moment.
1
z+1
=
1
z-1+2
=
1=2
1+
z+1
2
=
1
2
1
X
n=0
(-1)
n
z+1
2
n
From the full expressing, we obtain
2z+3
z+1
=
2z+3
2
1
X
n=0
(-1)
n
z+1
2
n
.
The radius of convergence can be found by1=R=lim sup
n!1
n
p
janj
. Therefore, the radius of conver-
gence is
R=1=lim sup
n!1
n
r
(-1)
n
1
2
n
=j2j=2
3.
X
n
p
z
n
,
X
z
n
n!
,
X
n!z
n
,
X
q
n
2
z
n
,
X
z
n!
wherejqj< 1.
For
P
n
p
z
n
, we can use the inverse of argument of the ratio test to determine the radius of convergence;
that is,
R=lim
n!1
n
p
(n+1)
p
=lim
n!1
n
p
(n+1)
p
=1
24
P
jfn(x)j6Mn<1
so
P
jfn(x)j
is absolutely convergent. Thus,
P
fn(x)
is pointwise
convergent tof(x). Let > 0be given andsn=
P
n
k=1
fk(x)benth partial sum. Letn > Nsuch that
jf(x) -snj=
1
X
k=1
fk(x) -
n
X
k=1
fk(x)
=
1
X
k=n+1
fk(x)
6
1
X
k=n+1
jfk(x)j
Since
P
Mk
converges to some limit, fornsufciently large,
P
1
k=n+1
Mk<
. SelectNsuch that this is
true. Then
jf(x) -snj6
1
X
k=n+1
jfk(x)j6
1
X
k=n+1
Mk<
Therefore,
P
fn(x)wherefn(x) =
x
n(1+nx
2
)
is uniformly convergent by the Weierstrass M-test.
6.
IfU=u1+u2+ ,V=v1+v2+ are convergent series, prove thatUV=u1v1+ (u1v2+u2v2) +
(u1v3+u2v2+u3v1) +
provided that at least one of the series is absolutely convergent. (It is easy
if both series are absolutely convergent. Try to rearrange the proof so economically that the absolute
convergence of the second series is not needed.)
2.2.4 Power Series
1. (1-z)
-m
,ma positive integer, in powers ofz.
The Binomial theorem states that(1+x)
n
=
P
1
k=0
n
k
x
k
. In our case, we have
(1-z)
-m
=
1
X
k=0
-m
k
(-z)
k
(2.6)
where
-m
k
= (-1)
k
m+k-1
k
. Then equation (2.6) can be written as
(1-z)
-m
=
1
X
k=0
m+k-1
k
z
k
=1+mz+
m(m+1)
2!
z
2
+ .
2.
2z+3
z+1
in powers ofz-1. What is the radius of convergence?
Let's just consider
1
z+1
for the moment.
1
z+1
=
1
z-1+2
=
1=2
1+
z+1
2
=
1
2
1
X
n=0
(-1)
n
z+1
2
n
From the full expressing, we obtain
2z+3
z+1
=
2z+3
2
1
X
n=0
(-1)
n
z+1
2
n
.
The radius of convergence can be found by1=R=lim sup
n!1
n
p
janj
. Therefore, the radius of conver-
gence is
R=1=lim sup
n!1
n
r
(-1)
n
1
2
n
=j2j=2
3.
X
n
p
z
n
,
X
z
n
n!
,
X
n!z
n
,
X
q
n
2
z
n
,
X
z
n!
wherejqj< 1.
For
P
n
p
z
n
, we can use the inverse of argument of the ratio test to determine the radius of convergence;
that is,
R=lim
n!1
n
p
(n+1)
p
=lim
n!1
n
p
(n+1)
p
=1
24
For
P
z
n
n!
, we can use the fact that the sum ise
zwhich is entire or the method used previously. Since
e
z
is entire, the radius of convergence isR=1.
R=lim
n!1
(n+1)!
n!
=lim
n!1
n!(n+1)
n!
=1
For
P
n!z
n
, we use the modied ratio test again.
R=lim
n!1
n!
(n+1)!
=0
For
P
q
n
2
z
n
, we will use the root test.
R=1=lim sup
n!1
n
p
jq
n
j
n
=1=lim sup
n!1
jqj
n
forjqj< 1,R=1, and forjqj> 1,R=0. For
P
z
n!
, we will use the root test.
R=1=lim sup
n!1
n
q
jz
(n-1)!
j
n
=1=lim sup
n!1
jzj
(n-1)!
Whenjzj< 1,R=1, and whenjzj> 1,R=0.
4.
P
anz
n
has a radius of convergenceR, what is the radius of convergence of
P
anz
2n
? of
P
a
2
nz
n
?
Since
P
anz
n
has a radius of convergenceR,
R=lim
n!1
an
an+1
.
For
P
anz
2n
=z
2
P
anz
n
, we have
jzj
2
lim
n!1
an
an+1
=jzj
2
R
so the radius of convergence is
p
R. For
P
a
2
nz
n
, we have
lim
n!1
an
an+1
2
=R
2
.
5.f(z) =
P
anz
n
, what is
P
n
3
anz
n
?
Let's write out the rst few terms of
X
n
3
anz
n
=a1z+8a2z
2
+27a3z
3
+64a4z
4
+
Let's consider the rst three derivatives off(z).
f
0
(z) =
X
nanz
n-1
zf
0
(z) =
X
nanz
n
=a1z+2a2z
2
+3a3z
3
+ (2.7)
f
00
(z) =
X
n(n-1)anz
n-2
z
2
f
00
(z) =
X
n(n-1)anz
n
=2a2z
2
+6a3z
3
+12a4z
4
+ (2.8)
f
000
(z) =
X
n(n-1)(n-2)anz
n-3
z
3
f
000
(z) =
X
n(n-1)(n-2)anz
n
=6a3z
3
+24a4z
4
+60a5z
5
+ (2.9)
If we add equations (2.7) to (2.9), we have
zf
0
(z) +z
2
f
00
(z) +z
3
f
000
(z) =a1z+4a2z
2
+15a3z
3
+ 6=
X
n
3
anz
n
However, consider3z
2
f
00
(z) =6a2z
2
+18a3z
3
+36a4z
4
+ . Then
zf
0
(z) +3z
2
f
00
(z) +z
3
f
000
(z) =a1z+8a2z
2
+27a3z
3
+64a4z
4
=
X
n
3
anz
n
.
25
P
z
n
n!
, we can use the fact that the sum ise
zwhich is entire or the method used previously. Since
e
z
is entire, the radius of convergence isR=1.
R=lim
n!1
(n+1)!
n!
=lim
n!1
n!(n+1)
n!
=1
For
P
n!z
n
, we use the modied ratio test again.
R=lim
n!1
n!
(n+1)!
=0
For
P
q
n
2
z
n
, we will use the root test.
R=1=lim sup
n!1
n
p
jq
n
j
n
=1=lim sup
n!1
jqj
n
forjqj< 1,R=1, and forjqj> 1,R=0. For
P
z
n!
, we will use the root test.
R=1=lim sup
n!1
n
q
jz
(n-1)!
j
n
=1=lim sup
n!1
jzj
(n-1)!
Whenjzj< 1,R=1, and whenjzj> 1,R=0.
4.
P
anz
n
has a radius of convergenceR, what is the radius of convergence of
P
anz
2n
? of
P
a
2
nz
n
?
Since
P
anz
n
has a radius of convergenceR,
R=lim
n!1
an
an+1
.
For
P
anz
2n
=z
2
P
anz
n
, we have
jzj
2
lim
n!1
an
an+1
=jzj
2
R
so the radius of convergence is
p
R. For
P
a
2
nz
n
, we have
lim
n!1
an
an+1
2
=R
2
.
5.f(z) =
P
anz
n
, what is
P
n
3
anz
n
?
Let's write out the rst few terms of
X
n
3
anz
n
=a1z+8a2z
2
+27a3z
3
+64a4z
4
+
Let's consider the rst three derivatives off(z).
f
0
(z) =
X
nanz
n-1
zf
0
(z) =
X
nanz
n
=a1z+2a2z
2
+3a3z
3
+ (2.7)
f
00
(z) =
X
n(n-1)anz
n-2
z
2
f
00
(z) =
X
n(n-1)anz
n
=2a2z
2
+6a3z
3
+12a4z
4
+ (2.8)
f
000
(z) =
X
n(n-1)(n-2)anz
n-3
z
3
f
000
(z) =
X
n(n-1)(n-2)anz
n
=6a3z
3
+24a4z
4
+60a5z
5
+ (2.9)
If we add equations (2.7) to (2.9), we have
zf
0
(z) +z
2
f
00
(z) +z
3
f
000
(z) =a1z+4a2z
2
+15a3z
3
+ 6=
X
n
3
anz
n
However, consider3z
2
f
00
(z) =6a2z
2
+18a3z
3
+36a4z
4
+ . Then
zf
0
(z) +3z
2
f
00
(z) +z
3
f
000
(z) =a1z+8a2z
2
+27a3z
3
+64a4z
4
=
X
n
3
anz
n
.
25
6.
If
P
anz
n
and
P
bnzn
have radii of convergenceR1andR2, show that the radii of convergence of
P
anbnz
n
is at leastR1R2.
Let > 0be given. Then there existsn > Nsuch that
janj
1=n
< 1=R1+, jbnj
1=n
< 1=R2+
since lim sup
n!1
janj
1=n
=1=R1sojanj
1=n
< 1=R1+and similarly forbn. Multiplying we obtain
janbnj
1=n
<
1
R1R2
+(1=R1+1=R2) +
2
Then
1
R
6
1
R1R2
)R1R26R
7. n!1janj=jan+1j=R, prove that
P
anz
n
has a radius of convergence ofR.
Let > 0be given. Supposejzj< R. Picksuch thatjzj< R-. Then for somen > N
an
an+1
-R
6R-
an
an+1
<
R- <
an
an+1
(2.10)
Forn > N, we can write
aN
an
=
aNaN+1 an-1
aN+1aN+2 an
=
aN
aN+1
aN+1
aN+2
an-1
an
(2.11)
Forn > N, we have that from equation (2.10),R- <
aN
aN+1
. Thus, we can equation (2.11) as
(R-)
n-N
<
aN
an
janj<
jaNj
(R-)
n-N
janz
n
j<jaNz
N
j
jzj
R-
n-N
Sincewas chosen such thatjzj< R-, we that
jzj
R-
< 1and
janz
n
j<jaNz
N
j
wherejaNz
N
j<1 since it is a convergent geometric series. Therefore,
P
anz
n
converges absolutely
with a radius of convergence ofR.
8. zis
1
X
n=0
z
1+z
n
convergent?
In order for series to converge lim sup
n!1
n
p
janj< 1. Then
lim sup
n!1
n
p
jz=(z+1)j
n
=
z
z+1
< 1
orjzj
2
<(1+z)(1+¯z) =1+2<fzg+jzj
2
so the series converges when
0 < 1+2<fzg.
26
If
P
anz
n
and
P
bnzn
have radii of convergenceR1andR2, show that the radii of convergence of
P
anbnz
n
is at leastR1R2.
Let > 0be given. Then there existsn > Nsuch that
janj
1=n
< 1=R1+, jbnj
1=n
< 1=R2+
since lim sup
n!1
janj
1=n
=1=R1sojanj
1=n
< 1=R1+and similarly forbn. Multiplying we obtain
janbnj
1=n
<
1
R1R2
+(1=R1+1=R2) +
2
Then
1
R
6
1
R1R2
)R1R26R
7. n!1janj=jan+1j=R, prove that
P
anz
n
has a radius of convergence ofR.
Let > 0be given. Supposejzj< R. Picksuch thatjzj< R-. Then for somen > N
an
an+1
-R
6R-
an
an+1
<
R- <
an
an+1
(2.10)
Forn > N, we can write
aN
an
=
aNaN+1 an-1
aN+1aN+2 an
=
aN
aN+1
aN+1
aN+2
an-1
an
(2.11)
Forn > N, we have that from equation (2.10),R- <
aN
aN+1
. Thus, we can equation (2.11) as
(R-)
n-N
<
aN
an
janj<
jaNj
(R-)
n-N
janz
n
j<jaNz
N
j
jzj
R-
n-N
Sincewas chosen such thatjzj< R-, we that
jzj
R-
< 1and
janz
n
j<jaNz
N
j
wherejaNz
N
j<1 since it is a convergent geometric series. Therefore,
P
anz
n
converges absolutely
with a radius of convergence ofR.
8. zis
1
X
n=0
z
1+z
n
convergent?
In order for series to converge lim sup
n!1
n
p
janj< 1. Then
lim sup
n!1
n
p
jz=(z+1)j
n
=
z
z+1
< 1
orjzj
2
<(1+z)(1+¯z) =1+2<fzg+jzj
2
so the series converges when
0 < 1+2<fzg.
26
9.
1
X
n=0
z
n
1+z
2n
.
Consider the following two equations:
j1j=j1+z
2n
-z
2n
j
6j1+z
2n
j+jz
2n
j
1-jz
2n
j6j1+z
2n
j (2.12)
jz
2n
j=j1-1+z
2n
j
6j1+z
2n
j+1
jz
2n
j-16j1+z
2n
j (2.13)
From equations (2.12) and (2.13), the triangle inequality, we have that
1-jz
2n
j
6j1+z
2n
j.
There exists anm > 1such that
jz
2n
j
m
6
1-jz
2n
j
.
By the root test,
lim sup
n!1
n
s
mjzj
n
jz
2
j
n
=lim sup
n!1
n
p
m
jzj
=
1
jzj
< 1
Whenjzj> 1, the convergence of the ratio test
1
jzj
< 1
leads tojzj> 1. Ifjzj< 1, then1=jzj> 1 where we
can write1=jzj=jz1j . Since the choice dummy variables is arbitrary,jzj< 1. In other words, the series
will converge whenjzj> 1orjzj< 1. Supposejzj=1. Then by the limit test,
lim
n!1
1
1
n
+1
-n
=
1
2
6=0;
therefore, the series diverges.
2.2.5 Abel's Limit Theorem
2.3 The Exponential and Trigonometric Functions
2.3.1 The Exponential
2.3.2 The Trigonometric Functions
1. (i), cos(i), and tan(1+i).
For sin(i), we can use the identity sin(z) =
e
iz
-e
-iz
2i
. Then
sin(i) =
e
-1
-e
1
2i
=i
e
1
-e
-1
2
=isinh(1).
Similarly, for cos(i), we have
cos(i) =
e
-1
+e
1
2
=
e
1
+e
-1
2
=cosh(1).
For tan(1+i), we can use the identity tan(z) = -i
e
iz
-e
-iz
e
iz
+e
-iz. Then
tan(1+i) = -i
e
i-1
-e
1-i
e
i-1
+e
1-i
= -itanh(i-1).
27
1
X
n=0
z
n
1+z
2n
.
Consider the following two equations:
j1j=j1+z
2n
-z
2n
j
6j1+z
2n
j+jz
2n
j
1-jz
2n
j6j1+z
2n
j (2.12)
jz
2n
j=j1-1+z
2n
j
6j1+z
2n
j+1
jz
2n
j-16j1+z
2n
j (2.13)
From equations (2.12) and (2.13), the triangle inequality, we have that
1-jz
2n
j
6j1+z
2n
j.
There exists anm > 1such that
jz
2n
j
m
6
1-jz
2n
j
.
By the root test,
lim sup
n!1
n
s
mjzj
n
jz
2
j
n
=lim sup
n!1
n
p
m
jzj
=
1
jzj
< 1
Whenjzj> 1, the convergence of the ratio test
1
jzj
< 1
leads tojzj> 1. Ifjzj< 1, then1=jzj> 1 where we
can write1=jzj=jz1j . Since the choice dummy variables is arbitrary,jzj< 1. In other words, the series
will converge whenjzj> 1orjzj< 1. Supposejzj=1. Then by the limit test,
lim
n!1
1
1
n
+1
-n
=
1
2
6=0;
therefore, the series diverges.
2.2.5 Abel's Limit Theorem
2.3 The Exponential and Trigonometric Functions
2.3.1 The Exponential
2.3.2 The Trigonometric Functions
1. (i), cos(i), and tan(1+i).
For sin(i), we can use the identity sin(z) =
e
iz
-e
-iz
2i
. Then
sin(i) =
e
-1
-e
1
2i
=i
e
1
-e
-1
2
=isinh(1).
Similarly, for cos(i), we have
cos(i) =
e
-1
+e
1
2
=
e
1
+e
-1
2
=cosh(1).
For tan(1+i), we can use the identity tan(z) = -i
e
iz
-e
-iz
e
iz
+e
-iz. Then
tan(1+i) = -i
e
i-1
-e
1-i
e
i-1
+e
1-i
= -itanh(i-1).
27
2.
The hyperbolic cosine and sine are dened ascosh(z) =
e
z
+e
-z
2
andsinh(z) =
e
z
-e
-z
2
. Express them
through cos(iz)and sin(iz). Derive the addition formulas, and formulas for cosh(2z)and sinh(2z).
For the rst part, we have
cos(iz) =
e
-z
+e
z
2
=cosh(z)
sin(iz) =
e
-z
-e
z
2i
=i
e
z
-e
-z
2
=isinh(z)
For cosh, we have that the addition formula is
cosh(a+b) =cos[i(a+b)]
=
e
a+b
+e
-(a+b)
2
=
2e
a+b
+2e
-(a+b)
4
=
e
a+b
+e
a-b
+e
b-a
+e
-(a+b)
+e
a+b
-e
a-b
-e
b-a
+e
-(a+b)
=4
=
e
a
+e
-a
2
e
b
+e
-b
2
+
e
a
-e
-a
2
e
b
-e
-b
2
=cosh(a)cosh(b) +sinh(a)sinh(b)
For sinh, we have that the addition formula is
sinh(a+b) = -isin[i(a+b)]
=
e
-(a+b)
-e
a+b
-2
=
2e
a+b
-2e
-(a+b)
4
=
e
a+b
+e
a-b
-e
b-a
-e
-(a+b)
+e
a+b
-e
a-b
+e
b-a
-e
-(a+b)
=4
=
e
a
-e
-a
2
e
b
+e
-b
2
+
e
a
+e
-a
2
e
b
-e
-b
2
=sinh(a)cosh(b) +cosh(a)sinh(b)
For the double angle formulas, recall thatcos(2z) =cos
2
(z) -sin
2
(z) =2cos
2
(z) -1=1-2sin
2
(z) and
sin(2z) =2sin(z)cos(z). Therefore, we have
cosh(2z) =cos(2iz)
=cos
2
(iz) -sin
2
(iz)
=
e
z
+e
-z
2
2
+
e
z
-e
-z
2
2
=cosh
2
(z) +sinh
2
(z)
cosh(2z) =2cos
2
(iz) -1
=2cosh
2
(z) -1
cosh(2z) =1-2sin
2
(iz)
=1-2sinh
2
(z)
sinh(2z) = -isin(2iz)
= -2isin(iz)cos(iz)
=2
e
z
-e
-z
2
e
z
+e
-z
2
=2sinh(z)cosh(z)
28
The hyperbolic cosine and sine are dened ascosh(z) =
e
z
+e
-z
2
andsinh(z) =
e
z
-e
-z
2
. Express them
through cos(iz)and sin(iz). Derive the addition formulas, and formulas for cosh(2z)and sinh(2z).
For the rst part, we have
cos(iz) =
e
-z
+e
z
2
=cosh(z)
sin(iz) =
e
-z
-e
z
2i
=i
e
z
-e
-z
2
=isinh(z)
For cosh, we have that the addition formula is
cosh(a+b) =cos[i(a+b)]
=
e
a+b
+e
-(a+b)
2
=
2e
a+b
+2e
-(a+b)
4
=
e
a+b
+e
a-b
+e
b-a
+e
-(a+b)
+e
a+b
-e
a-b
-e
b-a
+e
-(a+b)
=4
=
e
a
+e
-a
2
e
b
+e
-b
2
+
e
a
-e
-a
2
e
b
-e
-b
2
=cosh(a)cosh(b) +sinh(a)sinh(b)
For sinh, we have that the addition formula is
sinh(a+b) = -isin[i(a+b)]
=
e
-(a+b)
-e
a+b
-2
=
2e
a+b
-2e
-(a+b)
4
=
e
a+b
+e
a-b
-e
b-a
-e
-(a+b)
+e
a+b
-e
a-b
+e
b-a
-e
-(a+b)
=4
=
e
a
-e
-a
2
e
b
+e
-b
2
+
e
a
+e
-a
2
e
b
-e
-b
2
=sinh(a)cosh(b) +cosh(a)sinh(b)
For the double angle formulas, recall thatcos(2z) =cos
2
(z) -sin
2
(z) =2cos
2
(z) -1=1-2sin
2
(z) and
sin(2z) =2sin(z)cos(z). Therefore, we have
cosh(2z) =cos(2iz)
=cos
2
(iz) -sin
2
(iz)
=
e
z
+e
-z
2
2
+
e
z
-e
-z
2
2
=cosh
2
(z) +sinh
2
(z)
cosh(2z) =2cos
2
(iz) -1
=2cosh
2
(z) -1
cosh(2z) =1-2sin
2
(iz)
=1-2sinh
2
(z)
sinh(2z) = -isin(2iz)
= -2isin(iz)cos(iz)
=2
e
z
-e
-z
2
e
z
+e
-z
2
=2sinh(z)cosh(z)
28
3. (x+iy)and sin(x+iy)in real and imaginary parts.
cos(x+iy) =cos(x)cos(iy) -sin(x)sin(iy)
=cos(x)cosh(y) -isin(x)sinh(y)
sin(x+iy) =sin(x)cos(iy) +sin(iy)cos(x)
=sin(x)cosh(y) +isinh(y)cos(x)
4.
jcos(z)j
2
=sinh
2
(y) +cos
2
(x) =cosh
2
(y) -sin
2
(x) = (cosh(2y) +cos(2x))=2
and
jsin(z)j
2
=sinh
2
(y) +sin
2
(x) =cosh
2
(y) -cos
2
(x) = (cosh(2y) -cos(2x))=2.
For the identities, recall thatcosh
2
(z) -sinh
2
(z) =1 andcos
2
(z) +sin
2
(z) =1 . Then for the rst identity,
we have
jcos(z)j
2
=cos(z)cos(¯z)
=
cos(x)cosh(y) -isin(x)sinh(y)
cos(x)cosh(y) +isin(x)sinh(y)
=cos
2
(x)cosh
2
(y) +sin
2
(x)sinh
2
(y)
=cos
2
(x)(1+sinh
2
(y)) +sin
2
(x)sinh
2
(y)
=cos
2
(x) +sinh
2
(y)
jcos(z)j
2
=cos
2
(x)cosh
2
(y) +sin
2
(x)(cosh
2
(y) -1)
=cosh
2
(y) -sin
2
(x)
jcos(z)j
2
=
= (cosh(2y) +cos(2x))=2
jsin(z)j
2
=sin(z)sin(¯z)
=
sin(x)cosh(y) +isinh(y)cos(x)
sin(x)cosh(y) -isinh(y)cos(x)
=sin
2
(x)cosh
2
(y) +sinh
2
(y)cos
2
(x)
=sin
2
(x)(1+sinh
2
(y)) +sinh
2
(y)cos
2
(x)
=sin
2
(x) +sinh
2
(y)
jsin(z)j
2
=sin
2
(x)cosh
2
(y) + (cosh
2
(y) -1)cos
2
(x)
=cosh
2
(y) -cos
2
(x)
jsin(z)j
2
=
= (cosh(2y) -cos(2x))=2
2.3.3 Periodicity
2.3.4 The Logarithm
1.
For realy, show that every remainder in the series forcos(y) andsin(y) has the same sign as the leading
term (this generalizes the inequalities used in the periodicity proof).
The series for both cosine and sine are
cos(y) =
1
X
k=0
(-1)
k
y
2k
(2k)!
=1-
y
2
2!
+
y
4
4!
-
sin(y) =
1
X
k=0
(-1)
k
y
2k+1
(2k+1)!
=y-
y
3
3!
+
y
6
6!
-
29
cos(x+iy) =cos(x)cos(iy) -sin(x)sin(iy)
=cos(x)cosh(y) -isin(x)sinh(y)
sin(x+iy) =sin(x)cos(iy) +sin(iy)cos(x)
=sin(x)cosh(y) +isinh(y)cos(x)
4.
jcos(z)j
2
=sinh
2
(y) +cos
2
(x) =cosh
2
(y) -sin
2
(x) = (cosh(2y) +cos(2x))=2
and
jsin(z)j
2
=sinh
2
(y) +sin
2
(x) =cosh
2
(y) -cos
2
(x) = (cosh(2y) -cos(2x))=2.
For the identities, recall thatcosh
2
(z) -sinh
2
(z) =1 andcos
2
(z) +sin
2
(z) =1 . Then for the rst identity,
we have
jcos(z)j
2
=cos(z)cos(¯z)
=
cos(x)cosh(y) -isin(x)sinh(y)
cos(x)cosh(y) +isin(x)sinh(y)
=cos
2
(x)cosh
2
(y) +sin
2
(x)sinh
2
(y)
=cos
2
(x)(1+sinh
2
(y)) +sin
2
(x)sinh
2
(y)
=cos
2
(x) +sinh
2
(y)
jcos(z)j
2
=cos
2
(x)cosh
2
(y) +sin
2
(x)(cosh
2
(y) -1)
=cosh
2
(y) -sin
2
(x)
jcos(z)j
2
=
= (cosh(2y) +cos(2x))=2
jsin(z)j
2
=sin(z)sin(¯z)
=
sin(x)cosh(y) +isinh(y)cos(x)
sin(x)cosh(y) -isinh(y)cos(x)
=sin
2
(x)cosh
2
(y) +sinh
2
(y)cos
2
(x)
=sin
2
(x)(1+sinh
2
(y)) +sinh
2
(y)cos
2
(x)
=sin
2
(x) +sinh
2
(y)
jsin(z)j
2
=sin
2
(x)cosh
2
(y) + (cosh
2
(y) -1)cos
2
(x)
=cosh
2
(y) -cos
2
(x)
jsin(z)j
2
=
= (cosh(2y) -cos(2x))=2
2.3.3 Periodicity
2.3.4 The Logarithm
1.
For realy, show that every remainder in the series forcos(y) andsin(y) has the same sign as the leading
term (this generalizes the inequalities used in the periodicity proof).
The series for both cosine and sine are
cos(y) =
1
X
k=0
(-1)
k
y
2k
(2k)!
=1-
y
2
2!
+
y
4
4!
-
sin(y) =
1
X
k=0
(-1)
k
y
2k+1
(2k+1)!
=y-
y
3
3!
+
y
6
6!
-
29
We can write Taylor's formula asf(y) =Tn(y) +Rn(y)where
f(y) =
n
X
k=0
f
(k)
(0)
k!
y
k
+
1
k!
Z
y
0
(y-t)
k
f
(k+1)
(t)dt.
Now, we can write cosine and sine ofyas
cos(y) =
n
X
k=0
(-1)
k
y
2k
(2k)!
+
1
n!
Z
y
0
(y-t)
n
cos
n+1
(t)dt
sin(y) =
n-1
X
k=0
(-1)
k
y
2k+1
(2k+1)!
+
1
n!
Z
y
0
(y-t)
n
sin
n
(t)dt
For cosine and sine, letn=2mandn=2m-1, respectively. Then
cos(y) =
m
X
k=0
(-1)
k
y
2k
(2k)!
+
1
(2m)!
Z
y
0
(y-t)
2m
cos
2m+1
(t)dt
sin(y) =
m-1
X
k=0
(-1)
k
y
2k+1
(2k+1)!
+
1
(2m-1)!
Z
y
0
(y-t)
2m-1
sin
2m-1
(t)dt
2. 3 < < 2
p
3.
3. e
z
forz= -i=2,3i=4,2i=3.
e
-i=2
= -i
e
3i=4
= (-
p
2+i
p
2)=2
e
2i=3
= (-1+i
p
3)=2
4. zise
z
equal to2,-1,i,-i=2,-1-i,1+2i?
For all problems,k2Z.
e
z
=2 e
z
= -1
z=log(2) z=log(-1)
e
z
=i =logjij+i(arg(-1) +2k)
z=logjij+i(arg(i) +2k) = i(1+2k)
=
i
2
(1+4k) e
z
=
-i
2
e
z
= -1-i = -log(2) -
i
2
(1+4k)
z=logj-1-ij+i(arg(-1-i) +2k)e
z
=1+2i
=log(
p
2) -
3i
4
+2ki z =log(
p
5) +i(arctan(2) +2k)
=
log(2)
2
-
3i
4
+2ki =
log(5)
2
+i(arctan(2) +2k)
5. (e
z
).
Letz=x+iy. Then
exp(e
z
) =exp
e
x
(cos(y) +isin(y)
=exp(e
x
cos(y))exp(ie
x
sin(y))
=exp(e
x
cos(y))
cos(e
x
sin(y)) +isin(e
x
sin(y))
u(x,y) =exp(e
x
cos(y))cos(e
x
sin(y))
v(x,y) =exp(e
x
cos(y))sin(e
x
sin(y))
whereu(x,y)is the real andv(x,y)is the imaginary part of exp(e
z
).
30
f(y) =
n
X
k=0
f
(k)
(0)
k!
y
k
+
1
k!
Z
y
0
(y-t)
k
f
(k+1)
(t)dt.
Now, we can write cosine and sine ofyas
cos(y) =
n
X
k=0
(-1)
k
y
2k
(2k)!
+
1
n!
Z
y
0
(y-t)
n
cos
n+1
(t)dt
sin(y) =
n-1
X
k=0
(-1)
k
y
2k+1
(2k+1)!
+
1
n!
Z
y
0
(y-t)
n
sin
n
(t)dt
For cosine and sine, letn=2mandn=2m-1, respectively. Then
cos(y) =
m
X
k=0
(-1)
k
y
2k
(2k)!
+
1
(2m)!
Z
y
0
(y-t)
2m
cos
2m+1
(t)dt
sin(y) =
m-1
X
k=0
(-1)
k
y
2k+1
(2k+1)!
+
1
(2m-1)!
Z
y
0
(y-t)
2m-1
sin
2m-1
(t)dt
2. 3 < < 2
p
3.
3. e
z
forz= -i=2,3i=4,2i=3.
e
-i=2
= -i
e
3i=4
= (-
p
2+i
p
2)=2
e
2i=3
= (-1+i
p
3)=2
4. zise
z
equal to2,-1,i,-i=2,-1-i,1+2i?
For all problems,k2Z.
e
z
=2 e
z
= -1
z=log(2) z=log(-1)
e
z
=i =logjij+i(arg(-1) +2k)
z=logjij+i(arg(i) +2k) = i(1+2k)
=
i
2
(1+4k) e
z
=
-i
2
e
z
= -1-i = -log(2) -
i
2
(1+4k)
z=logj-1-ij+i(arg(-1-i) +2k)e
z
=1+2i
=log(
p
2) -
3i
4
+2ki z =log(
p
5) +i(arctan(2) +2k)
=
log(2)
2
-
3i
4
+2ki =
log(5)
2
+i(arctan(2) +2k)
5. (e
z
).
Letz=x+iy. Then
exp(e
z
) =exp
e
x
(cos(y) +isin(y)
=exp(e
x
cos(y))exp(ie
x
sin(y))
=exp(e
x
cos(y))
cos(e
x
sin(y)) +isin(e
x
sin(y))
u(x,y) =exp(e
x
cos(y))cos(e
x
sin(y))
v(x,y) =exp(e
x
cos(y))sin(e
x
sin(y))
whereu(x,y)is the real andv(x,y)is the imaginary part of exp(e
z
).
30
6. 2
i
,i
i
,(-1)
2i
.
For all problems,k2Z.
z=2
i
=exp
ilog(2)
=cos(log(2)) +isin(log(2))
z=i
i
=exp
ilog(i)
=exp[-(1+4k)=2]
z= (-1)
2i
=i
4i
= (i
i
)
4
=exp[-2(1+4k)]
7. z
z
.
Letz=x+iyandk2Z.
z
z
= (x+iy)
x+iy
=exp
(x+iy)log(x+iy)
=e
x=2log(x
2
+y
2
)-y(arctan(y=x)+2k)
h
cos
x(arctan(y=x) +2k) +y=2log(x
2
+y
2
)
+
isin
x(arctan(y=x) +2k) +y=2log(x
2
+y
2
)
i
Thus, the real part is
u(x,y) =e
x=2log(x
2
+y
2
)-y(arctan(y=x)+2k)
cos
x(arctan(y=x) +2k) +y=2log(x
2
+y
2
)
and the imaginary part is
v(x,y) =e
x=2log(x
2
+y
2
)-y(arctan(y=x)+2k)
sin
x(arctan(y=x) +2k) +y=2log(x
2
+y
2
)
8. (w)in terms of the logarithm.
Letarctan(w) =z . Thenw=tan(z) . Recall thattan(z) = -i
e
iz
-e
-iz
e
iz
+e
-iz
. Now, lete
2iz
=x . Then we have the
following
w= -i
x
2
-1
x
2
+1
which leads to
e
2iz
=
i-w
i+w
.
By taking the log, we can recoverz.
2iz=log(i-w) -log(i+w)
=log(i) +log(1+iw) -log(i) -log(1-iw)
z=
i
2
log(1-iw) -log(1+iw)
arctan(w) =z
=
i
2
log(1-iw) -log(1+iw)
9.
Show how to dene the "angles" in a triangle, bearing in mind that they should lie between0and.
With this denition, prove that the sum of the angles is.
31
i
,i
i
,(-1)
2i
.
For all problems,k2Z.
z=2
i
=exp
ilog(2)
=cos(log(2)) +isin(log(2))
z=i
i
=exp
ilog(i)
=exp[-(1+4k)=2]
z= (-1)
2i
=i
4i
= (i
i
)
4
=exp[-2(1+4k)]
7. z
z
.
Letz=x+iyandk2Z.
z
z
= (x+iy)
x+iy
=exp
(x+iy)log(x+iy)
=e
x=2log(x
2
+y
2
)-y(arctan(y=x)+2k)
h
cos
x(arctan(y=x) +2k) +y=2log(x
2
+y
2
)
+
isin
x(arctan(y=x) +2k) +y=2log(x
2
+y
2
)
i
Thus, the real part is
u(x,y) =e
x=2log(x
2
+y
2
)-y(arctan(y=x)+2k)
cos
x(arctan(y=x) +2k) +y=2log(x
2
+y
2
)
and the imaginary part is
v(x,y) =e
x=2log(x
2
+y
2
)-y(arctan(y=x)+2k)
sin
x(arctan(y=x) +2k) +y=2log(x
2
+y
2
)
8. (w)in terms of the logarithm.
Letarctan(w) =z . Thenw=tan(z) . Recall thattan(z) = -i
e
iz
-e
-iz
e
iz
+e
-iz
. Now, lete
2iz
=x . Then we have the
following
w= -i
x
2
-1
x
2
+1
which leads to
e
2iz
=
i-w
i+w
.
By taking the log, we can recoverz.
2iz=log(i-w) -log(i+w)
=log(i) +log(1+iw) -log(i) -log(1-iw)
z=
i
2
log(1-iw) -log(1+iw)
arctan(w) =z
=
i
2
log(1-iw) -log(1+iw)
9.
Show how to dene the "angles" in a triangle, bearing in mind that they should lie between0and.
With this denition, prove that the sum of the angles is.
31
10.
Show that the roots of the binomial equationz
n
=a are the vertices of a regular polygon (equal sides
and angles).
Letz=re
i
. Then
r
n
e
in
=ae
2ik
.
Therefore,r=a
1=n and=2ik=n . Sinceris just the radius, the roots will be located on a circle of
radiusratexp(2ik=n) fork2[0,n-1]. Since each root are angle multiplies about the origin, they will
benequally spaced points.nequally spaced points will form the vertices of a regularn-gon.
32
Show that the roots of the binomial equationz
n
=a are the vertices of a regular polygon (equal sides
and angles).
Letz=re
i
. Then
r
n
e
in
=ae
2ik
.
Therefore,r=a
1=n and=2ik=n . Sinceris just the radius, the roots will be located on a circle of
radiusratexp(2ik=n) fork2[0,n-1]. Since each root are angle multiplies about the origin, they will
benequally spaced points.nequally spaced points will form the vertices of a regularn-gon.
32
3 Analytic Functions as Mappings
3.1 Elementary Point Set Topology
3.1.1 Sets and Elements
3.1.2 Metric Spaces
1.
IfSis a metric space with distance functiond(x,y), show thatSwith the distance function(x,y) =
d(x
,y)=[1+d(x ,y)]is also a metric space. The latter space is bounded in the sense that all distances lie
under a xed bound.
Sinced(x,y)is a metric onS,d(x,y)satises
d(x,y)>0, and zero only whenx=y (3.1a)
d(x,y) =d(y,x) (3.1b)
d(x,z)6d(x,y) +d(y,z) (3.1c)
By equation (3.1a), forx=y,d(x,y) =0 so(x,y) =0=1=0 , and whenx6=y,d(x,y)> 0so(x,y)> 0
since a positive number divided by a positive number is positive. We have that(x,y)>0 and equal
zero if and only ifx=y. By equation (3.1b), we have
(x,y) =
d(x,y)
1+d(x,y)
=
d(y,x)
1+d(y,x)
=(y,x)
For the triangle inequality, we have
d(x,z)
1+d(x,z)
6
d(x,y)
1+d(x,y)
+
d(y,z)
1+d(y,z)
Let's multiple through by the product of all three denominators. After simplifying, we obtain
d(x,z)6d(x,y) +d(y,z) +2d(x,y)d(y,z) +d(x,y)d(y,z)d(x,z)
We have already shown thatd(x,y)>0 and zero if and only ifx=y. Ifx=y=z , the triangle inequality
is vacuously true. Whenx6=y6=z , the triangle inequality follows since each distance is positive and
equation (3.1c); that is,
(x,z)6(x,y) +(y,z).
2.
Suppose that there are given two distance functionsd(x,y)andd1(x,y)on the same spaceS. They are
said to be equivalent if they determine the same open sets. Show thatdandd1are equivalent if to
every > 0there exists a > 0such thatd(x,y)< impliesd1(x,y)< , and vice versa. Verify that this
condition is fullled in exercise1.
Let, > 0be given. We can write(x,y) =
d(x,y)
1+d(x,y)
=1-
1
1+d(x,y)
. We need to nd asuch that
wheneverd(x,y)< ,(x,y)< .
1-
1
1+d(x,y)
<
d(x,y)<
1-
33
3.1 Elementary Point Set Topology
3.1.1 Sets and Elements
3.1.2 Metric Spaces
1.
IfSis a metric space with distance functiond(x,y), show thatSwith the distance function(x,y) =
d(x
,y)=[1+d(x ,y)]is also a metric space. The latter space is bounded in the sense that all distances lie
under a xed bound.
Sinced(x,y)is a metric onS,d(x,y)satises
d(x,y)>0, and zero only whenx=y (3.1a)
d(x,y) =d(y,x) (3.1b)
d(x,z)6d(x,y) +d(y,z) (3.1c)
By equation (3.1a), forx=y,d(x,y) =0 so(x,y) =0=1=0 , and whenx6=y,d(x,y)> 0so(x,y)> 0
since a positive number divided by a positive number is positive. We have that(x,y)>0 and equal
zero if and only ifx=y. By equation (3.1b), we have
(x,y) =
d(x,y)
1+d(x,y)
=
d(y,x)
1+d(y,x)
=(y,x)
For the triangle inequality, we have
d(x,z)
1+d(x,z)
6
d(x,y)
1+d(x,y)
+
d(y,z)
1+d(y,z)
Let's multiple through by the product of all three denominators. After simplifying, we obtain
d(x,z)6d(x,y) +d(y,z) +2d(x,y)d(y,z) +d(x,y)d(y,z)d(x,z)
We have already shown thatd(x,y)>0 and zero if and only ifx=y. Ifx=y=z , the triangle inequality
is vacuously true. Whenx6=y6=z , the triangle inequality follows since each distance is positive and
equation (3.1c); that is,
(x,z)6(x,y) +(y,z).
2.
Suppose that there are given two distance functionsd(x,y)andd1(x,y)on the same spaceS. They are
said to be equivalent if they determine the same open sets. Show thatdandd1are equivalent if to
every > 0there exists a > 0such thatd(x,y)< impliesd1(x,y)< , and vice versa. Verify that this
condition is fullled in exercise1.
Let, > 0be given. We can write(x,y) =
d(x,y)
1+d(x,y)
=1-
1
1+d(x,y)
. We need to nd asuch that
wheneverd(x,y)< ,(x,y)< .
1-
1
1+d(x,y)
<
d(x,y)<
1-
33
Let=
1-
. For < 1, ifd(x,y)< =
1-
, then
(x,y) =1-
1
1+d(x,y)
< 1-
1
1+
1-
=.
If > 1,(x,y) =
d(x,y)
1+d(x,y)
< 1 <
. For the reverse implication, we need to nd asuch thatd(x,y)< .
Let=
1+
. For any > 0, if(x,y)< =
1+
, then
d(x,y)
1+d(x,y)
<
1+
d(x,y)(1+)< +d(x,y)
d(x,y)<
as was needed to be shown. Therefore,d(x,y)and(x,y)are equivalent metrics onS.
3. jz-z0j< isjz-z0j6.
4.Xis the set of complex numbers whose real and imaginary parts are rational, what is IntX,¯X,@X?
5.
It is sometimes typographically simpler to writeX
0forX. With this notation, how isX
0
-
0related to
X? Show thatX
-
0
-
0
-
0
-
0
=X
-
0
-
0
.
6. RorCis countable.
LetSbe a discrete set inRorC. Ifz2S, then for somei> 0, fori2Z,Ni
(zi) is thei-th neighborhood
ofzi. SinceSis discrete, there exists anfor eachisuch that the only point inNi
(zi) iszi. Letibe
this. Consider the following function
f(i) =
8
>
>
>
>
<
>
>
>
>
:
0, i=0
1, i=1
2i, i > 0
2(-i) +1,i < 0
We have putiin a one-to-one correspondence withZ
+
. Therefore,Sis countable.
7.
LetEbe a set. ThenE
0is the set of accumulation (limit) points. Ifziis a limit point,zi2E
0 . Now,zi
are limit points ofEas well. Thenfzig!zwherez2¯E. Therefore,z2E
0
soE
0
is closed.
3.1.3 Connectedness
1.
IfXS, show that the relatively open (closed) subsets ofXare precisely those sets that can be expressed
as the intersection ofXwith an open (closed) subsets ofS.
LetfUgbe the open sets ofSsuch that
S
U=S
. ThenX=X\
S
U=
S
(X\U) . LetfAng=
fX\Ug
. ThenAnis relatively open inXsinceAnbelongs to the topology ofX; that is, for eachn,
AnXandX=
S
n
An.
2.
For the rst implication,), suppose on the contrary that the union of two regions is a region and they
have no point in common. LetAandBbe these two nonempty regions. Since they share no point in
common,A\¯B=B\¯A=? . Therefore,AandBare separated so they cannot be a region. We have
reached a contradiction so if the union of two regions is a region, then they have a point in common.
For the nally implication, suppose they have a point in common and the union of two regions is not
a region. Since the union of two regions is not a region, the regions are separated. LetAandBbe
two nonempty separated regions. SinceAandBare separated,A\¯B=B\¯A=? ; therefore,Aand
Bcannot have a point in common. We have reached contradiction so if they have a point in common,
then the union of two regions is are a region.
34
1-
. For < 1, ifd(x,y)< =
1-
, then
(x,y) =1-
1
1+d(x,y)
< 1-
1
1+
1-
=.
If > 1,(x,y) =
d(x,y)
1+d(x,y)
< 1 <
. For the reverse implication, we need to nd asuch thatd(x,y)< .
Let=
1+
. For any > 0, if(x,y)< =
1+
, then
d(x,y)
1+d(x,y)
<
1+
d(x,y)(1+)< +d(x,y)
d(x,y)<
as was needed to be shown. Therefore,d(x,y)and(x,y)are equivalent metrics onS.
3. jz-z0j< isjz-z0j6.
4.Xis the set of complex numbers whose real and imaginary parts are rational, what is IntX,¯X,@X?
5.
It is sometimes typographically simpler to writeX
0forX. With this notation, how isX
0
-
0related to
X? Show thatX
-
0
-
0
-
0
-
0
=X
-
0
-
0
.
6. RorCis countable.
LetSbe a discrete set inRorC. Ifz2S, then for somei> 0, fori2Z,Ni
(zi) is thei-th neighborhood
ofzi. SinceSis discrete, there exists anfor eachisuch that the only point inNi
(zi) iszi. Letibe
this. Consider the following function
f(i) =
8
>
>
>
>
<
>
>
>
>
:
0, i=0
1, i=1
2i, i > 0
2(-i) +1,i < 0
We have putiin a one-to-one correspondence withZ
+
. Therefore,Sis countable.
7.
LetEbe a set. ThenE
0is the set of accumulation (limit) points. Ifziis a limit point,zi2E
0 . Now,zi
are limit points ofEas well. Thenfzig!zwherez2¯E. Therefore,z2E
0
soE
0
is closed.
3.1.3 Connectedness
1.
IfXS, show that the relatively open (closed) subsets ofXare precisely those sets that can be expressed
as the intersection ofXwith an open (closed) subsets ofS.
LetfUgbe the open sets ofSsuch that
S
U=S
. ThenX=X\
S
U=
S
(X\U) . LetfAng=
fX\Ug
. ThenAnis relatively open inXsinceAnbelongs to the topology ofX; that is, for eachn,
AnXandX=
S
n
An.
2.
For the rst implication,), suppose on the contrary that the union of two regions is a region and they
have no point in common. LetAandBbe these two nonempty regions. Since they share no point in
common,A\¯B=B\¯A=? . Therefore,AandBare separated so they cannot be a region. We have
reached a contradiction so if the union of two regions is a region, then they have a point in common.
For the nally implication, suppose they have a point in common and the union of two regions is not
a region. Since the union of two regions is not a region, the regions are separated. LetAandBbe
two nonempty separated regions. SinceAandBare separated,A\¯B=B\¯A=? ; therefore,Aand
Bcannot have a point in common. We have reached contradiction so if they have a point in common,
then the union of two regions is are a region.
34
3.
LetTbe a topological space such thatE,¯ET. LetEbe a connected set and suppose¯Eis separated;
that is,¯E=A[B whereA,Bare relatively open in¯E, nonempty, and disjoint sets. Then there exists
open setsU,VinTsuch thatA=U\¯E andB=V\¯E . Now,AU ,BV , andU,V6=? . Therefore,
U\E6=?6=V\E soU\E andV\E are nonempty, disjoint sets. ThenE=U[V soEis separated. We
have, thus, reached a contradiction and the closure of connected set is also connected.
4.
LetAbe the set of points(x,y)2R
2 withx=0,jyj61 , and letBbe the set withx > 0,y=sin(1=x) . Is
A[Bconnected?
This is known as the topologist's sine curve.0.20.40.60.81-1-0.50.51
Figure3.1: x2(0,1].
LetS=A[B . We claim that the closure ofBinR
2is¯B=S. Letx2S. Ifx2B, then take a constant
sequencefx,x,: : :g. Ifx2A , thenx= (0 ,y)forjyj61 or said another wayy=sin() for2[- ,].
We can writey=sin() asy=sin(+2k) fork2Z
+ . Letxk=1=(+2k)> 0 . Theny=sin(1=xk) .
Nowfxkg!0 whenk!1 . Then(xk,sin(1=xk)) = (xn ,y)!(0 ,y)2¯B sincejyj61 . Therefore,S¯B .
Letf(xn,yn)g2S such thatf(xn,yn)g!(x ,y)2R
2 . Thenlimn!1xn=x andlimn!1yn=y . From
the denition of the sets,x>0 andjyj61 sojyj=limn!1jynj61 . Ifx=0, then(0,y)2S since
jyj61 . Supposex > 0. Then there existm > Nsuch thatxm> 0 for allm > Nso(xn,yn)2B . Let
yn=sin(1=xn) since(xn,yn)2B . Notice that forz2(0 ,1),sin(1=zis continuous. Sincefxng!x and
yn=sin(1=xn), we have
y=lim
n!1
yn=lim
n!1
sin(1=xn) =sin(1=x).
Thus,(x,y)2¯BS so¯B=S. SinceA\¯B=A\S=A6=? ,Sis connected. However,Sis not path
connected. That is, being connected doesn't imply path connectedness.
SupposeSis path connected and there exists anf: [0,1]!S such thatf(0)2B andf(1)2A . SinceA
is path connected, supposef(1) = (0 ,1). Let=1=2 > 0. By continuity, for > 0,jf(t) - (0 ,1)j< 1=2
whenever1-6t61 . Sincefis continuous, the image off([1- ,1])is connected. Letf(1-) = (x ,y).
Consider the composite off: [1- ,1]!R
2 and its projection on thex-axis. Since both maps are
continuous as well as their composite, the image of the composite map is a connected subset ofR
1
which contains zero andx. Now zero is thex-coordinate off(1)andxthex-coordinate off(1-) . Since
R
1is convex, connected sets are intervals. Then the set ofx-coordinates forf(1-) isx02[0 ,x]. For
x02(0 ,x], there existst2[1- ,1]such thatf(t) = (x0 ,sin(1=x0)) . Ifx0=1=(2k-=2) fork1 , then
0 < x0< x . Now1=x0=(4k-1)=2 which is a2mutliple of-=2for allk. Therefore,sin(1=x0) = -1
so(x0,sin(1=x0)) = (1=(2k-=2) ,-1)for somet2[1- ,1]which lies within a distance of=1=2 of
(0,1). However, the distance between(1=(2k-=2) ,-1)and(0,1)for largekis greater than1which is
a contradiction. Thus,Scannot be path connected.
5.
LetEbe the set of points(x,y)2R
2 such that06x61 and eithery=0 ory=1=n for some positive
integern. What are the components ofE? Are they all closed? Are they relatively open? Verify thatE
is not locally connected.
35
LetTbe a topological space such thatE,¯ET. LetEbe a connected set and suppose¯Eis separated;
that is,¯E=A[B whereA,Bare relatively open in¯E, nonempty, and disjoint sets. Then there exists
open setsU,VinTsuch thatA=U\¯E andB=V\¯E . Now,AU ,BV , andU,V6=? . Therefore,
U\E6=?6=V\E soU\E andV\E are nonempty, disjoint sets. ThenE=U[V soEis separated. We
have, thus, reached a contradiction and the closure of connected set is also connected.
4.
LetAbe the set of points(x,y)2R
2 withx=0,jyj61 , and letBbe the set withx > 0,y=sin(1=x) . Is
A[Bconnected?
This is known as the topologist's sine curve.0.20.40.60.81-1-0.50.51
Figure3.1: x2(0,1].
LetS=A[B . We claim that the closure ofBinR
2is¯B=S. Letx2S. Ifx2B, then take a constant
sequencefx,x,: : :g. Ifx2A , thenx= (0 ,y)forjyj61 or said another wayy=sin() for2[- ,].
We can writey=sin() asy=sin(+2k) fork2Z
+ . Letxk=1=(+2k)> 0 . Theny=sin(1=xk) .
Nowfxkg!0 whenk!1 . Then(xk,sin(1=xk)) = (xn ,y)!(0 ,y)2¯B sincejyj61 . Therefore,S¯B .
Letf(xn,yn)g2S such thatf(xn,yn)g!(x ,y)2R
2 . Thenlimn!1xn=x andlimn!1yn=y . From
the denition of the sets,x>0 andjyj61 sojyj=limn!1jynj61 . Ifx=0, then(0,y)2S since
jyj61 . Supposex > 0. Then there existm > Nsuch thatxm> 0 for allm > Nso(xn,yn)2B . Let
yn=sin(1=xn) since(xn,yn)2B . Notice that forz2(0 ,1),sin(1=zis continuous. Sincefxng!x and
yn=sin(1=xn), we have
y=lim
n!1
yn=lim
n!1
sin(1=xn) =sin(1=x).
Thus,(x,y)2¯BS so¯B=S. SinceA\¯B=A\S=A6=? ,Sis connected. However,Sis not path
connected. That is, being connected doesn't imply path connectedness.
SupposeSis path connected and there exists anf: [0,1]!S such thatf(0)2B andf(1)2A . SinceA
is path connected, supposef(1) = (0 ,1). Let=1=2 > 0. By continuity, for > 0,jf(t) - (0 ,1)j< 1=2
whenever1-6t61 . Sincefis continuous, the image off([1- ,1])is connected. Letf(1-) = (x ,y).
Consider the composite off: [1- ,1]!R
2 and its projection on thex-axis. Since both maps are
continuous as well as their composite, the image of the composite map is a connected subset ofR
1
which contains zero andx. Now zero is thex-coordinate off(1)andxthex-coordinate off(1-) . Since
R
1is convex, connected sets are intervals. Then the set ofx-coordinates forf(1-) isx02[0 ,x]. For
x02(0 ,x], there existst2[1- ,1]such thatf(t) = (x0 ,sin(1=x0)) . Ifx0=1=(2k-=2) fork1 , then
0 < x0< x . Now1=x0=(4k-1)=2 which is a2mutliple of-=2for allk. Therefore,sin(1=x0) = -1
so(x0,sin(1=x0)) = (1=(2k-=2) ,-1)for somet2[1- ,1]which lies within a distance of=1=2 of
(0,1). However, the distance between(1=(2k-=2) ,-1)and(0,1)for largekis greater than1which is
a contradiction. Thus,Scannot be path connected.
5.
LetEbe the set of points(x,y)2R
2 such that06x61 and eithery=0 ory=1=n for some positive
integern. What are the components ofE? Are they all closed? Are they relatively open? Verify thatE
is not locally connected.
35
6. 3).
7.
A set is said to bediscreteif all its points are isolated. Show that a discrete set in a separable metric
space is countable.
3.1.4 Compactness
1.
Given an alternate proof of the fact that every bounded sequence of complex numbers has a convergent
subsequence (for instance by use of the limes inferior).
2.
Show that the Heine-Borel property can also be expressed in the following manner: Every collection of
closed sets with an empty intersection contains a nite subcollection with an empty intersection.
The statement above is equivalent to: A collectionFof closed subsets of a topological space(X,T)
has the nite intersection property if\F6=? for all nite subcollectionsFF . Show that(X,T),
a topological space, is compacy if and only if every family of closed setsFP(X) having the nite
intersection property satises\F6=?.
LetF=fF:2Agbe a collection of closed sets inX. Now
\
F=?()
[
F
c
=X.
Therefore, the set
S
F
c
is an open cover ofXsinceFis closed. If the intersection of the setfFn
g
is empty for a niten, then
S
F
c
n
is a nite subcover ofX. Then every open cover ofXhas a
nite subcover if and only if every collection of closed sets with an empty intersection has a nite
subcollection with an empty intersection. Thus,Xis compact if and only if every collection of closed
sets with the nite intersection property has a nonempty intersection.
3.
Since we are dealing with a set of real numbers, we are speaking of compact metric spaces. A subset
Eof a metric spaceXis compact if and only if every sequence inEhas convergent subsequence inE
(sequentially compact).
First, we will show)by contradiction. Letfxngbe a sequence inE. Supposefxngdoesn't have a
convergent subsequence inE. Then forx2E, there exists > 0such thatxn2N(x) for only nitely
manyn. ThenN(x) would be an open cover ofEwhich has no nite subcover. Therefore,Ecouldn't
be compact contradicting the premise. Thus, ifEis compact metric space, thenEis sequentialy compact.
In order to prove(, we need to prove that a sequentially compact set contains a countable dense
subset.
Lemma3.1.4.1:A sequentially compact set contains a countable dense subset (separable space).
LetAbe an innite sequentially compact set. SinceAis sequential compact,Ais bounded; otherwise, we
would have nonconvergent subsequences inA. Letfyngbe a dense sequence inA. Choosey1,y2,: : :,yn
offyng. Letn=sup
y2A
mink6nd(y ,yn)> 0. Letyn+1 be such thatd(yn+1 ,yk)>=2 fork=1,: : :,n.
Sincefynghas a convergent subsequence, for all > 0there existm,n2Z
+ such thatd(ym,yn)< .
Then
d(y,yn-1)< n-1=2 < d(ym,yn)< ()n-1< 2
Thus, ally2A is in2ofykfork < n. Since > 0and arbitrary,fyngis dense inAbecause every
nonempty open set contains at least one element of the sequence.
Now for(. LetFbe an open coverEand letfyngbe a dense sequence. Letr2Q and letGbe the
family of neighborhoods,Nr(yn) , that are contained inF. SinceQis countable,Gis countable. Let
x2E andx2F . ThenN(x)F for > 0. Sincefyngis dense inE, by lemma3.1.4.1,d(y,yn)< =2 for
somen. For allr2Q,d(yn,y)< r < -d(yn ,y). Thenx2Nr(yn)N(x)G . Sincex2Nr(yn)2G
andGis countable, we can nd a nite subcover ofG. Replace eachGbyFwhereGF for some
. Then this set ofFis a nite subcover. Thus, ifEis a sequentially compact metric space, thenEis
compact.
36
7.
A set is said to bediscreteif all its points are isolated. Show that a discrete set in a separable metric
space is countable.
3.1.4 Compactness
1.
Given an alternate proof of the fact that every bounded sequence of complex numbers has a convergent
subsequence (for instance by use of the limes inferior).
2.
Show that the Heine-Borel property can also be expressed in the following manner: Every collection of
closed sets with an empty intersection contains a nite subcollection with an empty intersection.
The statement above is equivalent to: A collectionFof closed subsets of a topological space(X,T)
has the nite intersection property if\F6=? for all nite subcollectionsFF . Show that(X,T),
a topological space, is compacy if and only if every family of closed setsFP(X) having the nite
intersection property satises\F6=?.
LetF=fF:2Agbe a collection of closed sets inX. Now
\
F=?()
[
F
c
=X.
Therefore, the set
S
F
c
is an open cover ofXsinceFis closed. If the intersection of the setfFn
g
is empty for a niten, then
S
F
c
n
is a nite subcover ofX. Then every open cover ofXhas a
nite subcover if and only if every collection of closed sets with an empty intersection has a nite
subcollection with an empty intersection. Thus,Xis compact if and only if every collection of closed
sets with the nite intersection property has a nonempty intersection.
3.
Since we are dealing with a set of real numbers, we are speaking of compact metric spaces. A subset
Eof a metric spaceXis compact if and only if every sequence inEhas convergent subsequence inE
(sequentially compact).
First, we will show)by contradiction. Letfxngbe a sequence inE. Supposefxngdoesn't have a
convergent subsequence inE. Then forx2E, there exists > 0such thatxn2N(x) for only nitely
manyn. ThenN(x) would be an open cover ofEwhich has no nite subcover. Therefore,Ecouldn't
be compact contradicting the premise. Thus, ifEis compact metric space, thenEis sequentialy compact.
In order to prove(, we need to prove that a sequentially compact set contains a countable dense
subset.
Lemma3.1.4.1:A sequentially compact set contains a countable dense subset (separable space).
LetAbe an innite sequentially compact set. SinceAis sequential compact,Ais bounded; otherwise, we
would have nonconvergent subsequences inA. Letfyngbe a dense sequence inA. Choosey1,y2,: : :,yn
offyng. Letn=sup
y2A
mink6nd(y ,yn)> 0. Letyn+1 be such thatd(yn+1 ,yk)>=2 fork=1,: : :,n.
Sincefynghas a convergent subsequence, for all > 0there existm,n2Z
+ such thatd(ym,yn)< .
Then
d(y,yn-1)< n-1=2 < d(ym,yn)< ()n-1< 2
Thus, ally2A is in2ofykfork < n. Since > 0and arbitrary,fyngis dense inAbecause every
nonempty open set contains at least one element of the sequence.
Now for(. LetFbe an open coverEand letfyngbe a dense sequence. Letr2Q and letGbe the
family of neighborhoods,Nr(yn) , that are contained inF. SinceQis countable,Gis countable. Let
x2E andx2F . ThenN(x)F for > 0. Sincefyngis dense inE, by lemma3.1.4.1,d(y,yn)< =2 for
somen. For allr2Q,d(yn,y)< r < -d(yn ,y). Thenx2Nr(yn)N(x)G . Sincex2Nr(yn)2G
andGis countable, we can nd a nite subcover ofG. Replace eachGbyFwhereGF for some
. Then this set ofFis a nite subcover. Thus, ifEis a sequentially compact metric space, thenEis
compact.
36
4.E1E2 is a decreasing sequence of nonempty compact sets, then the intersection
T
1
1
Enis not
empty (Cantor's lemma). Show by example that this need not be true if the sets are merely closed.
Consider the topological spaceR
1
. LetEn=fn2Z
>0
: [n,1)g. ThenEnEn+1 . Since(-1,n)is
open inR
1
,[n,1) = (-1,n)
c
is closed. The innite intersection ofEnis
1
\
n=1
En=
1
\
n=1
[n,1) =?.
Thus, the statement isn't true if we consider only closed sets.
5.
LetSbe the set of all sequencesx=fxng of real numbers such that only a nite number of thexnare
6=0. Dened(x,y) =maxjxn-ynj . Is the space complete? Show that the-neighborhoods are not
totally bounded.
3.1.5 Continuous Functions
1. jzj< 1onto the whole plane.
LetX=fz2C:jzj< 1g . Then we need to nd a function,f, such thatf:X!C . Consider the function
tan(z). Thenz2(-=2,=2)7!f(z)2(-1,1). Letf(z) =tan(z=2). Thenz2(-1,1)7!f(z)2(-1,1).
2.
Prove that a subset of the real line which is topologically equivalent to an open interval is an open
interval. (Consider the effect of removing a point.)
3.
Prove that every continuous one-to-one mapping of a compact space is topological. (Show that closed
sets are mapped on closed sets.)
4.
LetXandYbe compact sets in a complete metric space. Prove that there existsx2X and ay2Y such
thatd(x,y)is a minimum.
5.
sin(x),xsin(x),xsin(x
2
),
p
jxjsin(x)?
3.1.6 Topological Spaces
3.2 Conformality
37
T
1
1
Enis not
empty (Cantor's lemma). Show by example that this need not be true if the sets are merely closed.
Consider the topological spaceR
1
. LetEn=fn2Z
>0
: [n,1)g. ThenEnEn+1 . Since(-1,n)is
open inR
1
,[n,1) = (-1,n)
c
is closed. The innite intersection ofEnis
1
\
n=1
En=
1
\
n=1
[n,1) =?.
Thus, the statement isn't true if we consider only closed sets.
5.
LetSbe the set of all sequencesx=fxng of real numbers such that only a nite number of thexnare
6=0. Dened(x,y) =maxjxn-ynj . Is the space complete? Show that the-neighborhoods are not
totally bounded.
3.1.5 Continuous Functions
1. jzj< 1onto the whole plane.
LetX=fz2C:jzj< 1g . Then we need to nd a function,f, such thatf:X!C . Consider the function
tan(z). Thenz2(-=2,=2)7!f(z)2(-1,1). Letf(z) =tan(z=2). Thenz2(-1,1)7!f(z)2(-1,1).
2.
Prove that a subset of the real line which is topologically equivalent to an open interval is an open
interval. (Consider the effect of removing a point.)
3.
Prove that every continuous one-to-one mapping of a compact space is topological. (Show that closed
sets are mapped on closed sets.)
4.
LetXandYbe compact sets in a complete metric space. Prove that there existsx2X and ay2Y such
thatd(x,y)is a minimum.
5.
sin(x),xsin(x),xsin(x
2
),
p
jxjsin(x)?
3.1.6 Topological Spaces
3.2 Conformality
37
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