Class 11 Chapter 5 Complex Numbers and Quadratic Equations Lecture_1.pdf

#class_11 #JEE #complex_numbers
Class 11 Chapter 5 Complex Numbers and Quadratic Equations Lecture 1
#iota #modulus #sum #difference #multiplication #conjugate
#class_11
#class_11_math
#class_11_maths
#Mathematics_Online_Lectures
#Maths_Learning_Centre_Jalandhar

1. Dr. Pranav Sharma
Maths Learning Centre. Jalandhar.

2. Complex Numbers and Quadratic Equations

IMAGINARY NUMBERS: if the square of a given number is negative then such a
number is called an imaginary number.

vu, V=2, etc, are imaginary numbers.

we denote \—T by the Greek letter Lota "Y, which is transliterated as “2.

Thus, V—4 = 2i, V—9 = 3i and V-5 = iv5, eto.

$3139973UI[UOSINLUIYILN O) /LOIIANINOA

Maths Learning Centre, Jalandhar https://sites.google.com/view/mathslearningcentre

3. POWERS OF i

y

xi=(-1)xi=-i;
(-1) x (-1) =1

i

Thus, we have i = 1,1 = 4,2 = 1,8
Let us consider i”, where n is a positive integer and n > 4.
On dividing n by 4, let the quotient be m and the remainder ber. Then,
n=4m+r, where 0 <r < 4.

in = ¡Mmtr — ¡tm y jr = a xi =i. [it=1]

$3139973UI[UOSINLUIYILN O) /LOIIANINOA

Maths Learning Centre, Jalandhar https://sites.google.com/view/mathslearningcentre

4. (i) 198 = ¡22442 — a)” xi=i=-1 [it=]]

Ba nus a 2 2 u

(P= en) -1. 14% = 1 mad? =-11
2 ¡2

22 998 _ 1 A E; A 21000 _ (;4 250 _

Gi) i > == 3=-1 [00 = (it) =1

4 L { ¡4 —

(iv) i AE [=1]

WW = 12243 = (4)? x B Lx (DE 1. [8 =1]

$3139973UI[UOSINLUIYILN O) /LOIIANINOA

Maths Learning Centre, Jalandhar https://sites.google.com/view/mathslearningcentre

5. s 3 alas Mal
M?S PRES y y = Y ALT X

(745 =i+ Ci = 0.

(VD) FEES EE I = M1 + TF FE)
=i"(1+i-1-i) = (fx 0) = 0. [2 =-1and8 =- il

(ix) ¿107 E j112 ste i117 di i122 = 4071 = iD + ¿10 de is)
=f (Lot titi)
=07(14+14824+8)14=1,=1000 2 = 1]
= 0744 i-1-0=(07x0)=0.

$3139973UI[UOSINLUIYILN O) /LOIIANINOA

Maths Learning Centre, Jalandhar https://sites.google.com/view/mathslearningcentre

6. (x) ar a4 0tx (141k) sa 4 ota otf =

=(1+D4-D04= (1-2) = 41 -(-1)}* = 24 = 16.

Ano 2
Show that{i + E) } = —4.
ı
5
we have 123 = i4*5+3) = (10 x 8 = 1 x (ED = —i Li = 1 and BP =-i]

NS | TE = 5 el a 3

() =p pe * pa 2=3= il = ind = 1
129?

{ee +(3) } = (ic DD = 42-41) =

Maths Learning Centre, Jalandhar https://sites.google.com/view/mathslearningcentre

$3139973UI[UOSINLUIYILN O) /LOIIANINOA

7. O
simplif 020 (Ei) (4 vi 4 + 5-9 3/-16

o 20 (4) -(-2xi)xı
(ti) aV-4+5Y-9 = 31-16 = (4 x 2i) + (5x 30) + (8 X 4i)
= (8i+ 15i - 12i) = (11)i.

4n+3

$3139973UI[UOSINLUIYILN O) /LOIIANINOA

Show that (-V=1) =i, wheren is a positive integer.
4n+3
(=D = (93 = (y x (D = (D x (-B) = xd =k
Maths Learning Centre, Jalandhar https://sites.google.com/view/mathslearningcentre

8. Show that the sum (1+ 1? + it +--+ 2") is 0 when nis odd andt whenn is

even.

LeS=14+P4 1404127

This is clearly a GP having (n + 1) terms with a= Land r = ran

all- +) 1x (1 = (2 (1-2 (1-0)
(ine ntré, alander 2) Math Legs Cen 2

2a — 1) = 0, when n is odd

S=

za +1) = 1, when nis even.

For any two real numbers a and b, the result Va x Vb = Vab is true only when at

least one of the given numbers is either zero or positive.
Thus, V-2 x V-3 = /(-2) x (23) = V6 is wrong.
In fact, V—2 x V-3 = (iv2)(iv3) = 12 x V6 = -V6.

Maths Learning Centre, Jalandhar https://sites.google.com/view/mathslearningcentre

$3139973UI[UOSINLUIYILN O) /LOIIANINOA

9. explain the fallacy: 1 = GX) = VI x V=1 = JED) X CD SVT = 1.
we know that for any real numbers a and b, Va x Vb = Vab is true only when at
least one of a ana b is either O or positive.

=1 x V-1 # J(-1) xX CD.

(0, V=25 x V=49 = (Si) (7i) = (35 x #7) = 35 x (+1) =-35.

(it) V=36 x V16 = (6i) x 4 = 24i.

$3139973UI[UOSINLUIYILN O) /LOIIANINOA

Maths Learning Centre, Jalandhar https://sites.google.com/view/mathslearningcentre

10. Find the value of Di (+ i+) (where, i =V=1)
43. 5 =

Dieters) =) e+ Yet = G40) + (2 +0) = 1-1
ni net n=1

Find thevalue of Daeg im (where, i = V—1).
nlis divisible by 4, Vn > 4.
eg LM = 9 + 104 104. oF times = 97 (6)
oe oar o a + rt = 104 EL 12 4 18! +97 Thom ea (01
SPA PA E 97 111 1-097 95401

$3139973UI[UOSINLUIYILN O) /LOIIANINOA

Maths Learning Centre, Jalandhar https://sites.google.com/view/mathslearningcentre

11. Show that the polynomial xP + x44+4 + x4rt? + x4843 is divisible by x? + x? +
x +1, where p, g,7T,S EN.
Let f(x) = x4P + xt4+1 À aar+2 ae x4s+3
and x8 + x2 + x +1 = (x? + 1)(x +1) = (x + (x — D (x + 1), where i = V-1
Now, f (i) =P + (441 + rt 4 i443 = 141418244 = 0 [sum of four
consecutive powers of iis zerol

fo 5 (Dr ES (yt + (SS) lila E (EE)

=1+(-0'+ (i? + (-)3 =1-i-1+i=0
and f(—1). = DP EDAD E

=1-1+1-1=0

Hence, by division theorem, f(x) is divisible by x? +x? + x +1.

$3139973UI[UOSINLUIYILN O) /LOIIANINOA

Maths Learning Centre, Jalandhar https://sites.google.com/view/mathslearningcentre

12. COMPLEX NUMBERS The numbers of the form (a + ib), where a and b are real
numbers andi = \—1, are known as complex numbers.
The set of all complex numbers is denoted by C.
C = {(a+ ib): a,b € R}.
Each of the numbers (5 + Bi), (—3 + V2i) and E - : i) is a complex number.

For a complex number, z = (a + ib) , we have
a =real part of Z, written as Re (z)
and b = imaginary part ofz, written as Im (z).

(1) fz = (5 + 91) then Re (z) = 5 and Im (z) = 9.
(ú) ¡pz = (31) then Re (2) == ana Im (2) = -3.

(i) pz = (-7 +51) then Re (2) = -7 and Im (7) =}.

8

$3139973UI[UOSINLUIYILN O) /LOIIANINOA

Maths Learning Centre, Jalandhar https://sites.google.com/view/mathslearningcentre

13. PURELY REAL AND PURELY IMAGINARY NUMBERS

A complex number z is said to be

() purely real, if Im (z) = 0,

(it) purely imaginary, if Re (z) =0.

Thus, each of the numbers 2, —7, V3 is purely real.

And, each of the numbers 2i, (v3i), E i) is purely imaginary.

CONJUGATE OF A COMPLEX NUMBER
Conjugate of a complex number Z = (a + ib) is defined as,Z = (a—ib).

(0) (3 + 85) = (3-81) (i) (-6 — 28) = (-6 + 21) (it) -3 = -3.

$3139973UI[UOSINLUIYILN O) /LOIIANINOA

Maths Learning Centre, Jalandhar https://sites.google.com/view/mathslearningcentre

14. Representation of conjugate

$
Imaginary axis =
Pe) 8
o
, ©
D 3
=
¡0 a ®
> Real =
0 0 E
b axis =
3
SM
+
QZ 8
2 ó
=
5
o
=
®
3
=
=
oO
n
Maths Learning Centre, Jalandhar https://sites.google.com/view/mathslearningcentre

15. MODULUS OF A COMPLEX NUMBER
Modulus of a complex number Z = (a+ ib) , denoted by |z\, és defined as

|z| = Ya? + b2.
() Hz = (2 +31) then |z| = /22 + 32 = V13.
(i) tfz = (-5 - 4i) then |z| = (52 + 4? = VAL.

EQUALITY OF COMPLEX NUMBERS
tf Zy = 01 + iby and zz = a2 + ib, then Z1 = 22 S 41 = Az and by = bp.

if2y + (3x —y)i = (5 —2i), find the values of x andy.
equating the real and imaginary parts, we get 2y + Ge —H)i = (5-21)
2y = Sand 3x-y = 72 ey =Sand3x-3=-2eyasandx as

He x=tandy=2
nee, = ga y

$3139973UI[UOSINLUIYILN O) /LOIIANINOA

Maths Learning Centre, Jalandhar https://sites.google.com/view/mathslearningcentre

16. SUM AND DIFFERENCE OF COMPLEX NUMBERS
(fay = (aı + iby) andz, = (a, + ib) then we define.
( 21 + 22 = (ay + az) + ¿(b, + bz)

(2) 2 — 22 = (a, — az) + i(by — bp).

(1) £21 = (3 + 58) and 22 = (-5 + 21) then
Z1 +22 = {34+ (-5)} +5 +2}= (-2+70)
and Zi — Z2 = {3—(-5)}+i(5 — 2) = (8 +30.

(ti) tf zı = (-6 — Zi) and 22 = (—3 — 5i) then
Z1 +22 = {(-6) + (-3)} + i{(-2) + (-5)} = (-9- 7H)
and Za — Z2 = {-6 — (-3)} + i{(-2) - (-5)} = (-3 +30.

$3139973UI[UOSINLUIYILN O) /LOIIANINOA

Maths Learning Centre, Jalandhar https://sites.google.com/view/mathslearningcentre

17. (t) Re (zy +22) = Re (21) + Re (zz) and
Im (zı +22) = Im (zı) + Im (2).

(it) Re (zy — 22) = Re (z1) — Re (zz) and
Im (zı — Z2) = Im (z1) — Im(z2) .

$3139973UI[UOSINLUIYILN O) /LOIIANINOA

Maths Learning Centre, Jalandhar https://sites.google.com/view/mathslearningcentre

18. PROPERTIES OF ADDITION OF COMPLEX NUMBERS
() CLOSURE PROPERTY: The sum of two complex numbers is always à complex
number.
Let za = (ay + iby) and z, = (az + ib,) be amy two complex numbers.
Then, Z1 + Zz = (ay + iby) + (az + ib,)
(ay + az) + ¿(by + bz) , which is a complex number.
Thus, tf Z1 and zz are any two complex numbers then (z1 + z2) is also a complex
number.
(i) COMMUTATIVE LAW: For any two complex numbers Zı and Z2, prove that
Z1 +22 = 22 +21.
Let zı = (a, + iby) and zZ, = (az + ib), where ay, Az, by, bz are real numbers.
Z1 +22 = (a, + iby) + (a, + ib)

= (ay + az) + i(by + b2)

= (az + ay) + ¿(b, + by) [oy commutative Law of addition in R]
= (a, + ib2) + (ay + iby) = zZ, +24.

Thus, Z4 + Zz = Zz + 24 for all 24, 22 E C.
Hence, addition of complex numbers is commutative.

Maths Learning Centre, Jalandhar https://sites.google.com/view/mathslearningcentre

$3139973UI[UOSINLUIYILN O) /LOIIANINOA

19. (tit) ASSOCIATIVE LAW: For any complex numbers 24, Z2 ana Z3, prove that
(Z1 +22) + 23 = Z1 + (22 + 23).
Let Z = (ay + iby), 22 = (az + ib) and z3 = (az + ibz) , where ay, Az, Az and
bi, bz, b3 are all real numbers.
(21 + 22) + 23 = {(aı + iby) + (az + ib2)} + (az + ibz)

= {(aı + az) + ¿(by + b2)} + (az + ibs)

= ((a, + az) + az) + i{(bı + bz) + b3)

= [ay + (a, + a3)) + ify + (b2 + b3}

Toy assoctative Law of addition in R]
= (a, + iby) + {(az + a3) + ¿(b, + b3))
= (a, + iby) + {(a2 + ibz) + (az + ib3))
= Z1 + (22 +23).

(21 + 22) + 23 = Z + (22 +23) for all z1, 22, Z3 E C. Hence, addition of complex
numbers is associative.

$3139973UI[UOSINLUIYILN O) /LOIIANINOA

Maths Learning Centre, Jalandhar https://sites.google.com/view/mathslearningcentre

20. (iv) EXISTENCE OF ADDITIVE IDENTITY

For any complex number Zz, prove that Z+0=0+2=2.

Let z = (a + ib) and we may write, 0 = (0 +10). Then,
z+0=(a+ib)+(0+10) = (a+0)+i(b +0) = (a+ ib)

and 0 + z = (0 + i0) + (a+ib) =(04+a)+i(0+b)= (a+ ib).z +0 = 0 +

z = 2 for all values of x € C.

Thus, 0 ts the additive identity for complex numbers.

(v) EXISTENCE OF ADDITIVE INVERSE
For every complex number, prove that z + (—z) = (-z)+z=0.
Let z = (a+ ib) . Then, (—z) = a) + i(-b).
z+(-z) = (a+ ib) + {(-a) + i(-b)} = {a+ (-a)} + i{b + (-b)} = (0+ i0).
Similarly, (-2) +z = (0+ i0) = 0.
Hence, z+(-2) =(-2) +z= 0.
Thus, every complex number z has (=2) as tts additive inverse.

$3139973UI[UOSINLUIYILN O) /LOIIANINOA

Maths Learning Centre, Jalandhar https://sites.google.com/view/mathslearningcentre

21. MULTIPLICATION OF COMPLEX NUMBERS
Let zy = (aı + iby) and z, = (az + ib2) . Then, we define
Z122 = (a, + iby) (az + ib,) = (ayaz — by bz) + i(aybz + bıa,).
Z1Z2 = {Re (21): Re (zz) — Im (2) - Im (z2)}
+i{ Re (zı): Im (zz) + Re (zz) - Im (z:1)}.
() Let zy = (3 + 28) and z, = (5 + 4). Then,
2122 =((3X5)-(2x 4)} +4UBXHN+EQX5)}
= (15 — 8) + ¿(12 + 10) = (7+ 22i).
(it) Let zy = (-2 + 38) and z2 = (7 — 51). Then,
2422 = {@2)*% 7 =3 x (5) FHC2YX (57 +3 x7}
= (-144+15)+i(10+ 21) = (1 + 31i).

$3139973UI[UOSINLUIYILN O) /LOIIANINOA

Maths Learning Centre, Jalandhar https://sites.google.com/view/mathslearningcentre

22. PROPERTIES OF MULTIPLICATION OF COMPLEX NUMBERS
() CLOSURE PROPERTY The product of two complex numbers is always a complex
number.
Let zı = (ay + iby) and z, = (az + ib,) be any two complex numbers. Then, ay,
az and by, ba are real numbers.

Z1Z2 = (ay + iby)(az + ib2)
= (a, a, — by bz) + i(aybz + by az) , which is a complex number. Thus, the product
of two complex numbers is always a complex number.

(tt) COMMUTATIVE LAW: For any two complex numbers Zı and Z2, prove that
Z122 = 222.
Let zı = (a, + ibi)andz, = (az + ib2) . Then,
2122 = (ay + iby)(a, + ibz) = (asa, — byb2) + i(a1b2 + by a2)
= (aa, — b2b1) + i(bza1 + azb,) Loy commutativity of multiplication on R]
= (az + ib,)(aı + iby) = 7274.

$3139973UI[UOSINLUIYILN O) /LOIIANINOA

Maths Learning Centre, Jalandhar https://sites.google.com/view/mathslearningcentre

23. (iii) ASSOCIATIVE LAW: For any three complex numbers 24,22 and 23, prove that
(Z122)23 = Z1(2223).
Let zı = (ay + iby), 22 = (az + ibz) and z3 = (az + ibz) be any three complex
numbers. Then 44, 42, az and by, ba, bz are all real numbers.
(Z122)Z3 = {(aı + ibı)(a, + ibz)}(a3 + ibs)
= {(a1az — by bz) + i(a,b, + byaz)}(az + ib3)
= {(a1az — bıb,)az — (a1b, + b,a,)b3)
+if(aza, — b1b2)b3 + (a1b2 + b1a,)a3)
= (a, (azaz — by b3) — by(azb3 + ba3)}
+i(a,(azb3 + a3b2) + bi(azaz — byb3)}
= (a, + ibyY(azaz — b2b3) + i(azbz + a3bz)}
= (as + iby){(az2 + iby)(az + ib3)) = z1(2223) .
Thus, (2122)23 = 21 (2223) for all z1, 22, 23 EC.
Hence, multiplication of complex numbers is associative.

$3139973UI[UOSINLUIYILN O) /LOIIANINOA

Maths Learning Centre, Jalandhar https://sites.google.com/view/mathslearningcentre

24. (iv) EXISTENCE OF MULTIPLICATIVE IDENTITY
The complex number (1 + 10) is the multiplicative identity inc.
Let z= (a+ ib). Then,
zx1=(a+ib)(1+i0) = {(a x 1) — (bx 0)} + i{(a x 0) + (bx 1)}
= (a+ ib) =z.
Similarly, 1X 2=2
Thus, (2x1) = (1 z) =z forallze C.
Hence, the complex number 1 = (1 + i0) is the multiplicative identity.

(V) EXISTENCE OF MULTIPLICATIVE INVERSE Let z = (a+ ib) + 0. Then,
care 1 2 1 the ine Le Eh)
z (a+ib) (a+ib) (a-ib) (a*+b?)
Clearly, z X zl=zlxz=1
Thus, every z = (a+ ib) has its multiplicative inverse, given by

1_ (a-ib) _ =

zZ

$3139973UI[UOSINLUIYILN O) /LOIIANINOA

z 71 =- = = —
z (a?+b2) |z/
zz = (2/7
Maths Learning Centre, Jalandhar https://sites.google.com/view/mathslearningcentre

25. Remember: 271 = a
Z

z= (a+ ib) >Z = (a- ib) and |z|? = (a? +b”).

: u PA (a — ib)
z=(a+tib)>z SE @+b)

(vi) DISTRIBUTIVE LAWS: For complex numbers 24, Z2, Z3, prove that
Zi: (22 +23) = ZıZ2 + 2123. (24 +22): Z3 = 2423 + 2223.
Leza = (ay + iby), 22 = (az + ibz) and 23 = (az + ibs) be any three complex
numbers. Then, Ay, Az, Az and by, bz, bs are real numbers.
Z1 * (22 + Z3) = (ay + ib,)[(a + ib) + (az + ibz)]
= (ay +ib1)[(az + az) + i(b2 + b3)]
= {a;(az + a3) — by (bz + b3)) + i{a1(b2 + b3) + by (az + a3)}
= {(a1az — bıbz) + i(aybz + b1az)} + {(a1a3 — by b3) + i(a1b3 + b1a3)}
= Z122 + 2123:

Thus, 24(Z2 + Z3) = 2122 + 2123 for all Z1, Zz, 23 € C.
Similarly, we caw prove that (21 + 22)Z3 = 2423 + 2223.

Maths Learning Centre, Jalandhar https://sites.google.com/view/mathslearningcentre

$3139973UI[UOSINLUIYILN O) /LOIIANINOA

26. The complex numbers do not possess the property of order,
(a + ib) > or < (c + id) is not defined .

A real number a can be written as a + i - 0. Therefore, every
real number can be considered as a complex number, whose imaginary part is zero
NCWcICQCRCC

tf Z1 and Za are complex numbers, then zi + 23 = 0 does not imply zı = Z2 = 0.
For example, 214 = 1 + iand z2 = 1—idHere, Z1 # 0,22 #0

But

wi+2=(1+07+0-0% =14+24+2i414+7?-21=2427

Il
N
|
N
Il
o

However, tf product of two complex numbers ts zero, then at Least one of them must be
Zero.

$3139973UI[UOSINLUIYILN O) /LOIIANINOA

Maths Learning Centre, Jalandhar https://sites.google.com/view/mathslearningcentre

27. DIVISION OF TWO COMPLEX NUMBERS

<
Let Z1 and Zz be complex numbers such that zz + 0. Then, a =2Z1: A = 21237. 2
=
Find, when zı = (6 + 3i) and zz = (3-1). =
2 ©
z =
We have = Su. 3
_1 _ 22 _@)_ (8+) _ Bt

Now, 22° = 1,12 32 Gn 10° =
Zi a . B+) (6+30(3+1) =;
— = 24:25 >= (64+3i)- = eamingcentre; st 10)
Z 1° ( ET 10 3
15+15i 15(1+i) 3(14+0 =
= + = % [9]
10 10 2 à
=
5
o
=
®
3
=
5
oO
a

Maths Learning Centre, Jalandhar https://sites.google.com/view/mathslearningcentre

28. SOME IDENTITIES ON COMPLEX NUMBERS
For any complex numbers Zı and 22, prove that:
() (24 + 22)? = 29 + 23 + 22422
(1) (z1 — 22)? = 25 + 23 - 2212,
(ii) (zi = 23) = (21 + 22)(Zı — 22)
(iv) (24 +22)? = 23 +23 + 32122(Zı + 22)
Y (1 - 22)” = 24 - 23 - 32122(21 — 22)
() (z1 + 22)? = (21 + 22)(Zı + 22)
= (21 + 22)z1 + (21 + 22)22 [by distributive Law]
= 27 + 2221 + 2122 +25 [by distributive Law]
= 22 + 22422 + 22 [2224 = 2122]
= zt + = + 22122.
(za +22)? = 23 + 78 + 22420.

$3139973UI[UOSINLUIYILN O) /LOIIANINOA

Maths Learning Centre, Jalandhar https://sites.google.com/view/mathslearningcentre

29. (1) Gr = 22)? = (21 = 22) (@1 = 22)
= (21 — Z2)Z1 — (Zı — Z2)Z2 [by distributive Law]
= 2} - 2224 - 2422 + 23 [by distributive Law]
= 29 — 22127 + 23 [2221 = 2122]
= 24 + 23 — 22425.
(21-22? = 23 + 23 — 22422.

(it) RAS = (2, + 2)(24 — 22)
= (24 + 22)21 — (21 + 22)22 Loy distributive Law]
= Zi + 2221 — 2422 — z3 [by distributive Law]

= (zi — z3) = LHS.[z2z, = 2422]

(24 = 23) = (41 +22) = 22).

$3139973UI[UOSINLUIYILN O) /LOIIANINOA

Maths Learning Centre, Jalandhar https://sites.google.com/view/mathslearningcentre

30. (1) (a +22)9 = (y+ 22)? (21 +22)

= (24 + 23 + 2212,)(Z1 + 22) Iexpanding (21 + 22)?1

= (23 + iz + 22122)Z1 Al (23 a 2 = a 22422)Z2 Toy distributive Law]
= 2 + 2321 ee 22225 + Zz se Zz == 22423

= 23 + 23+ 32222 + 32122 = 23 + 2} + 32122 (zı +22).

(Za +22)-= 25 +23 + 32122(21 + 22).

(v) LHS = (zı — 22)?

= (21 2221 - 22) = (25 - 22122 +25) (21 — 22)
= (27 — 2212, + 22 me E — 22122 + 25)22

= zt = 22222 + zizi - zz; + 22123 = 2

= Zi — 3242, + 37173 — z3 = RHS.

This may be written as (24 — 22)° = 24 — 23 — 32122(21 — 22).

$3139973UI[UOSINLUIYILN O) /LOIIANINOA

Maths Learning Centre, Jalandhar https://sites.google.com/view/mathslearningcentre

31. Simplify

(0) 3(6 + 61) + i(6 + 61) = 18 + 181 + Gi + 61? = 18 + 24i — 6 = 12 + 241.

(i) 1-H -(-3+6i) =1-i+3-6i=4—- Ti.

Maths Learning Centre, Jalandhar https://sites.google.com/view/mathslearningcentre

$3139973UI[UOSINLUIYILN O) /LOIIANINOA

32. Express (=V3 + V=2)(2V3 = i) in the form (a+ ib):

we have (—V3 + V—2)(2v3 — i) = (-V3 + iV2)(2V3 — i)
= —6 + iv3 + 2V6i- v2i?
= (-6 + V2) + V3(1 + 2v2)i.

Express each of the following in the form (a + ib):
© (3 + V=5)(3 — V=5) = (3+ v5i)@ - vBi) = (3? - (v51)”)
= (9- 512) = (9+5) = 14 +01.

(i) (2 + V-3)(-3 + 2V-3) = (-2 + v3i)(-3 + 2431)
= (6 - 4/31 - 3V3i+ 62) = —7V3i = 0 — 7V3i.

$3139973UI[UOSINLUIYILN O) /LOIIANINOA

(it) (2 +30? = (-2)? + GH? +2 x (22) x 3i
= (44 98? - 12i) = (4-9=12i) = (-5-12i).

Maths Learning Centre, Jalandhar https://sites.google.com/view/mathslearningcentre

33. Prove that (1 — i)* =-4,
(4-4 = (1-8? x (1-1)? = (14+ # - 2) x (1+ # -2i)
CEQ 4 = 4 x (1) ea? 244]

Express (2 + 303 in the form (a + ib).
we know that (24 + 22)? = 25 + z +3z,22(21 + 22).

(2.4. 303 = 23 + (303 4+3.x2 x 3i x (24,31) = 8 +278 + 361+ 547
= (8 —27i + 36i — 54) = (-46 + 91)

$3139973UI[UOSINLUIYILN O) /LOIIANINOA

Maths Learning Centre, Jalandhar https://sites.google.com/view/mathslearningcentre

34. Express (3 — Si)? in the form (a + ib) .
We know that (zy — 22)? = zi — 23 — 32122(zı — 22).

(3 — 50° = 33 — (503 — 3 x3 x 5ix (3 — 5H = 27 — 12513 — 135i + 2251?
= (27 + 125i — 135i — 225) = (-198 — 101).

ıfz = (V2-V-3), find Re (2), Im (2),z and |z].
z= v2 -iv3.
Re (z) = v2, Im (z) = -v3, Z= (v2 + iv3)
and 1212 = {(v2)° + (-v3)'} = @+3) =5 > 121 = vB.

$3139973UI[UOSINLUIYILN O) /LOIIANINOA

Maths Learning Centre, Jalandhar https://sites.google.com/view/mathslearningcentre

35. write down the modulus of.
() Let z = 449-3. Then, z = 4 + iv3.
æ
Iz1? = {(4)? + (v3) } = (16 +3) = 19 > [zl = 19.

(i) Let z = 2 —5i. Then,
lal? = {2? + (-5)?} = (4 + 25) = 29 > |7] = V29.

(ii) Let z = 0 — i. Then,
Il? = {07 + (-1)?} = (0+ 1) =1> [7] =V1=1.

(iv) Let z = (-1 + 31)?. Then,
z= (-1) + (31)? +2 x (—1) x 3i = (-8- 61).
11? = (58)? + (-6)?} = (64 + 36) = 100 = 121 = V100 = 10.

$3139973UI[UOSINLUIYILN O) /LOIIANINOA

Maths Learning Centre, Jalandhar https://sites.google.com/view/mathslearningcentre

36. Write down the conjugate of each of the following:
(1) Letz = (—5 + i) . Then, Zz = (—5 — i)

(ti) Let z = —6 — iV3. Then, Z = (—6 — iv3).

(ti) Lez= B =-i=0-i Then, z=i

(iv) Letz = (44 50? = (4)? + (50? +2x4x5i
= (16 — 25 + 408) = (-9 + 401).
Z = (—9 — 408).

$3139973UI[UOSINLUIYILN O) /LOIIANINOA

Maths Learning Centre, Jalandhar https://sites.google.com/view/mathslearningcentre

37. Find the multiplicative inverse of each of the following:
(i) Lez = (V5 +3i). then, Z=(V5- 31) and

2_ 2 E _ 1-2 _ 5-3i)_(v5_3,
Iz? = (v5) + 32 = (5 +9) = 14. esa = (8-41).

(it) Leez=4-=3i Then, Z=(4-3i) =(4+ 30) and
1212 = {(4)? + (—3)?} = (16 +9) = 25.

= z 4+3i 4 3.
z1= 7 =! 2= (5 xi).
Izl 25 25

(ile w= BES)? (NHL — 6i) = HO HIS 6) = (28 61):
Z = (-8 — 6i) = (-8 + 6i)
and |z|? = {(-8)? + (-6)2} = (64 + 36) = 100.

= E = SHS 6 )=(2 3 i)
e ak? 100 — 100 * 100! ~ 25 ' 50! :

$3139973UI[UOSINLUIYILN O) /LOIIANINOA

Maths Learning Centre, Jalandhar https://sites.google.com/view/mathslearningcentre

38. (ty) Let z = (0 — i). Then,

Z=(-f =iand a?=/®+-1%=1. wt=Zatai

Express each of the following in the form (a + ib) :

»Maiths Laark Garde Ar (CE) _ Gdns bio tante
On am an an icon 2 2 — (ae)
(6) (-1+ van)” y cen, O A

m= 2155 (-1-v3i) (-1+V3i)(-1-v3i) (-1)2-(v3i)”

(13 _ C1 ee

(1334) 4 a 4

u SAVED (BZ) ANZ) | (SV ZO (14 vai) = Eat
(ii) 1-Y2i (1-v2i) = (1+v2i) (1-2i2) = (1 + 221).

Maths Learning Centre, Jalandhar https://sites.google.com/view/mathslearningcentre

$3139973UI[UOSINLUIYILN O) /LOIIANINOA

39. (Y) ( La Bn) Eo L (étre) Ey 2 445), (6-20)

1+2i 1+3i (1+2i)(1-i) 1+3i (3+i) (1431)
_ (4+5i)(3-2i) _ (12+10)+(15-8)i _ (22+7i)
~ (+)(0+3) ~~ (3-3)+(91+) 108 i
er (7i2+22i) =7+22i _ e view i) =] E 7 Cer i)
1012 -10 10 10 10 5

Reduce & = =) to the form (a + ib) and hence find its modulus.

OS ati IN a t0?+ G0? _ xix Mail,
Lez = a i “aan a 27
Thus, z=0+2i> |z| = 02 +2? = V4=2.
Hence, Z= 0+ 2iand |z| = 2.
Maths Learning Centre, Jalandhar https://sites.google.com/view/mathslearningcentre

$3139973UI[UOSINLUIYILN O) /LOIIANINOA

40. (a + ib) then show that (a? + b?) = 1.

vie TH _ (140) =o) =(4 i i)
va “Viel Je VE drum

del fi? A à
onan (bs 5)=19 a? +b?) =1.
la+ ib? =(=5) +(5) = +2 ( )
Hence, (a? + pb?) = 1.

- RR ay, (ata
Find the least positive integerm for which (5) =1.

1+i) (Ati) Ati) at)? _ (1+r2i+i?) 2i_ 5
we have G = 2 Csiro, Gitfhdh@-@ — 2Maths earn
1+i\™
( ) =1=i"=1
1-1

And, we know that 4 is the least positive integer such that it = 1 and therefore) m =
4

$3139973UI[UOSINLUIYILN O) /LOIIANINOA

Maths Learning Centre, Jalandhar https://sites.google.com/view/mathslearningcentre

41. Separate E —{i into real and imaginary parts and hence find its modulus.
_ (3H _ (340) _ Gti) _ GD +i)
ern (=A) u a) @-) (2-D 7 (2+)

> G+D(2+) (6+i2)+45i __ 5+Si u
~ 2-O(2+i) A-2) W--D}

al re

Hence, z= (1+ i) and |z| = V2.

=(1+1.

V5+12i+V5-1 a P m a
Reduce f era si) © the form (a + ib) and hence find its conjugate.
we have
CoVS¥12i+V5=12i _ (V5+12i+V5-12i) © (VS+12i4V5-121)
~ ¥5+12i-V5-12i (v5+12i-V5-12i) © (V5+12i+V5-12i)
WS REYS-12) _ (5+12+6-12+2,/6)2- 1202
o (5+120=(5-120 244
— 10+2V25+144 _ 10+2V169 _ (10+2x13) _ 36 _3_3_1_ Si _ _3;
u 24i 24i 24 MH id ii 22 2”
c = = .
Hence, Z = (0-32) and Z = (0-21) = (0+2i).
à 2 2
Maths Learning Centre, Jalandhar https://sites.google.com/view/mathslearningcentre

$3139973UI[UOSINLUIYILN O) /LOIIANINOA

42. (f(a + iy)/3 = (a+ ib) then prove that:
+7 = 4(a? - b?) (a) 2-7 = -2(a? + b?)
we have (x + int = (a+ 2 >(x+iy) = (a+ ib)?

> (x+iy) = a? + ¿Bb? + 3iab(a + ib)

= a? — ib? + 3a?bi — 3ab? = (a? — 3ab?) + i(3a?b — b?)

>x=.a? — 3ab? andy = 3a?b — b? Ion equating real and imaginary parts
separately]
> = = (a? - 3b?) and a = (3a? - b?)
> (4 x E - 1) and (= Er a +22).
Hence, (t) F425 4(a? pm. b?) and (ti) -- == —2(a? + HT

$3139973UI[UOSINLUIYILN O) /LOIIANINOA

Maths Learning Centre, Jalandhar https://sites.google.com/view/mathslearningcentre

43. Lez be a complex number such thatz #1 and \2\ = 1. Then, prove that (=) is
purely imaginary. What will be your conclusion when z = 1?
Let z = (x + iy) be the given complex number such that z + 1 and |z| = 1. Then,
zl=1=> {25 x2+y=1x2+y2-1=0. (1)
= 2 ED _ xtiy-1 _ @-Driy Dd (DH)
2+1) (241) xtiy+l (x+1)tiy O (x+1)-iy {@+1)2+y2}
_ (@2-14y92)+{@&+1)y-@-Dy) _ (a+y2-1)+@yi _ { 2yi
u {@+122+y2} DY lar?
which is purely imaginary.
Particular Case When z = 1. ln this case, z=1=>x+iy=1
=>(x-1)+iy=0>%*-1=0,jy=0>%x=1,y=0.
z-1 _ (x+iy)-1 _ (x-1)+iy _ (1-1)+ix0 _0

using O1

210 GANA ADAM Y ARDOZ 0.
z-1,
Thus, z=1> za purely real.
Maths Learning Centre, Jalandhar https://sites.google.com/view/mathslearningcentre

$3139973UI[UOSINLUIYILN O) /LOIIANINOA

44. Find the real value of @ for which (
We have

) és purely real.

qe sin = = G+2isin 6) „ (1+2i sin 6) _ (3+2isin 8)(1+2i sin 0)
1-2isin@/ ~ (4—2isin@) (1+2isin6) — (1-412sin28)
_ (8-4sin?6)+i(6 sin 0+2 sin 8) _ (3-4sin?@)+i(8 sin 0)

(1+4sin20) (1+4sin20)
8 sin 9

(1+4sin20) ”
This happens only when 8sind=0 2 sind=0e0d0=n7,neEN.
Hence, the required value of O is ni, where n EN.

+2i sin O

—2isin O.

Now, ( ) will be purely real only when,

$3139973UI[UOSINLUIYILN O) /LOIIANINOA

Maths Learning Centre, Jalandhar https://sites.google.com/view/mathslearningcentre

45. Show that \1 — iX = 2° has no nonzero integral solution.
x x x
[1 —i[* = 2* = (2) = 2* 5 22 =2* 54 =1522=1=2°
22
> 2 =0>x=0.

Thus, x = 0 is the only solution of the given equation.
Hence, the given equation has no nonzero integral solution.

$3139973UI[UOSINLUIYILN O) /LOIIANINOA

Maths Learning Centre, Jalandhar https://sites.google.com/view/mathslearningcentre

46. Solve forx and y:

(i) The given equation is (3x — 7) + 2iy = —5y + (5 + x)i.

Equating the real parts and the imaginary parts of the given equation separately,
we get

3x-7=-5yand2y=5+x >3x+5y=7andx—-2y=-5
>x=-1andy=2 Ion solving 3x + 5y = 7,x —2y =-—51.

Hence, x = —1 and y = 2.

(it) The given equation is (x + iy)(2—3i) =4+i
> (2x +3y) + i(2y — 3x) = 4+i = 2x + 3y = 4 and 2y —-3x = 1
Tequating real parts and imaginary parts separately]
> 2x +3y = 4 and —3x + 2y = 1
5 14 4 _ af _
> x= and y = 7, Lon solving 2x + 3y = 4, —-3x + 2y = 11.

5 14, + /
Hence, X = 3 and y = 3s the required solution.

$3139973UI[UOSINLUIYILN O) /LOIIANINOA

Maths Learning Centre, Jalandhar https://sites.google.com/view/mathslearningcentre

47. sait
y1_,_&@D G-ù , 0-D Es _
we have à it si an rote em”
EUA Dp Ge DSH) 1; de-D-10-D ya Delon) Ce
(9-12) 9-2) 10 10 o
> 3(x- 1) - i(x- 1) + 3(y — 1) + ¿(y — 1) = 10i
=> (3x-3+3y=3)+i(y=,1-—x+1) = 101
> (3x +3y — 6) + i(y — x) = 101
2 3(x+y—2) = 0 and y — x = 10
Tequating real parts and the imaginary parts separately]
=>xt+y—-2=0andy—x=10 >x+y=20d-x+y=10
= x = —4 and y = 6 [on solving x + y = 2, —x + y = 101.
Hence, X = 4 and y = 6 are the required values.

Find the E values of x and y for which E

Maths Learning Centre, Jalandhar https://sites.google.com/view/mathslearningcentre

$3139973UI[UOSINLUIYILN O) /LOIIANINOA

48. Find the complex number 2 for which |z +1] =z+2(1+i).
Let the required complex number bez = (x + iy) . Then,

Iz+1]=z2+2(1+0> |@wtiy)+1|=(«+iy) +2(11+H
> J@+D2?+y2 = (x +2) + iy +2)
> J@41? +y? = (42) and y+2=0

[equating real parts and imaginary parts separately]

y =-2 and. (e+ 1)* (2)? =@+ 2)
=>y=-2Zand Vx? +2x4+5 = (x42)
>y=-2 and (x? +2x +5) = (x + 2)?
D x2+2x+5 = x? +4x +4 and y = —-2
> 2x\= Landy =-2=x= Sand y =-2,

Hence, the required complex number is Z = G — 2i) .

$3139973UI[UOSINLUIYILN O) /LOIIANINOA

Maths Learning Centre, Jalandhar https://sites.google.com/view/mathslearningcentre

49. Solve the equation \z\ + z= (2 + à) for complex value of 2.
Let z = (x + iy) . Then,

J|+2=24i> [xt+iyl+(xtiy)=24i> [Ft ++ iy=(2+0
Sy + y2 +x=2andy=1

Tequating real parts and imaginary parts separately]
= y= Land Vx2+12#x=23 y = Land Vx2+1=(2-x)
= y = Land x2 +1 = (2-x)2 = y = 1 and x? +1 = x2 —- 4x +4
34x = Bandy = 15 x= and y = 1.

3 2, ir 3 4
Hence, Z = E + i) ts the desired solution.

$3139973UI[UOSINLUIYILN O) /LOIIANINOA

Maths Learning Centre, Jalandhar https://sites.google.com/view/mathslearningcentre

50. Solve the equation 2+V2\z +1] + i = 0 for complex value of 2.
Let the required complex number be z = x + iy. Then, z+V2]z2+1|+i=0
> (x+iy) + V2](x+iy) +11+i=0

> x+v2{V@+D? + y?) + + Di=0

2 x + V2(JG+D7+y7)=0 and y +1=0
Tequating real parts and imaginary parts separately on both sides]

> y = Land V2{/G + D7+ CD} = o)
y=-1mndV2-/x2+2x +2 = (-x)
=> y =—Land 2 (x? +2x+2)=x?
>y=-1andx?+4x+4=0> (x +2)? = 0 and y = -1
3x+2= 0d y =-15 yx =-Zandy =—1.
Hence, the required complex number is z = (-2 —i).

$3139973UI[UOSINLUIYILN O) /LOIIANINOA

Maths Learning Centre, Jalandhar https://sites.google.com/view/mathslearningcentre

51. (fa = 3 + 2i, prove that 2? — 62+ 13 = 0 and hence deduce that
323 —13z7+9z+65=0.

Z2=342i5 2-3=2i> (2-3)? = 47 > 2? -—6z4+13=0. ()

Thus, 22 — 6z+13 =0.

Now, 32? — 132? + 97 + 65 = 3z(z? — 6z + 13) + 5(2? — 6z + 13)

=(32x0)+(5x0)=0 [using (1.

Hence, 32° — 132? + 9z + 65 = O.

ifz = —5 +31, find the value of (z* + 923 + 262? — 147 + 8).
We havez = —5 + 3i => (2 + 5) = 31
> (z+5)? = 97 > 2? +10z+34=0. (i)
Now, z* + 9z3 + 267? — 14z + 8
= 22 (2? + 10z + 34) — z(z? + 10z + 34) + 2(z? + 10z + 34) — 60
= (22 x 0) — (2x0) + (2 x 0) — 60 = -60 [using (1.

$3139973UI[UOSINLUIYILN O) /LOIIANINOA

Maths Learning Centre, Jalandhar https://sites.google.com/view/mathslearningcentre

52. (fa = 2 +i, prove that 23 +327 — 9z + 8 = (1+14i).
we have
z=2+i>z-2=i>(2-2?=i2=>2-424+5=0. (i)
23432? -92+8=z(2?°-42+5)+7(z?- 4245) +14z2- 27
= (zx 0) + (7 x 0) + 14z - 27 = (14z — 27) [using (1
= 14(2+i) —27 = (1+14i).
Hence, 2° +32? — 97 +8 = (1 + 14i).

$3139973UI[UOSINLUIYILN O) /LOIIANINOA

Maths Learning Centre, Jalandhar https://sites.google.com/view/mathslearningcentre