not.pdf , freshman Mathematics math1011

This note used for every first year university students.

1. WalleTilahun(MSc)walle T
MIZAN TEPI UNIVERSITY
COLLEGE OF NATURAL AND COMPUTATIONAL SCIENCES
Fresh Man Mathematics for Natural Sciences
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2. WalleTilahun(MSc)walle T
Math 1011 teaching material for section I
Proofs and Examples are not included under this
Material please try to refer your module!
Set By:Walle Tilahun(MSc)
Email: [email protected]
If there is error reply me!
January 22, 2024
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3. WalleTilahun(MSc)walle T
UNIT 1:The Proposition
Definition of Proposition
In logic and philosophy, a proposition is a statement that declares a fact or
asserts a claim that can be either true or false. It is the building block of
logical reasoning and argumentation.
Proposition
Human activities are the primary cause of global warming.
Example 1: The increase in greenhouse gas emissions from industrial
processes and transportation.
Example 2: Deforestation and land use changes leading to higher
levels of carbon dioxide in the atmosphere.
Non-Proposition Examples
Please turn off the lights when you leave the room. (Command)
I enjoy hiking in the mountains. (Hobby)
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4. WalleTilahun(MSc)walle T
Logical Connectives and Rules
Conjunction (∧)
Rule: The conjunctionp∧qis true if and only if bothpandqare true.
Disjunction (∨)
Rule: The disjunctionp∨qis true if at least one ofporqis true.
Negation (¬)
Rule: The negation¬pis true if and only ifpis false.
Implication (→)
Rule: The implicationp→qis false only whenpis true andqis false.
Biconditional (↔)
Rule: The biconditionalp↔qis true if and only ifpandqhave the
same truth value.
Suppose the truth values ofp,q,andrareT,F,andF
respectively,then the truth value of(¬p∨q)→[q↔(p∧r)]is
. . . . . .
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5. WalleTilahun(MSc)walle T
What is the Converse , Inverse, and Contrapositive of
p→q?
I. If a rectangle ”R” is a square, then ”R” is a rhombusp→q.
Converse: If ”R” is a rhombus, then ”R” is a square q→p)
Inverse: If ”R” is not a square, then ”R” is not a rhombus¬p→ ¬q.
Contrapositive: If ”R” is not a rhombus, then ”R” is not a square
¬q→ ¬p.
II. If today is Monday, then tomorrow is Tuesdayp→q.
Converse: If tomorrow is Tuesday, then today is Monday q→p)
Inverse: If today is not Monday, then tomorrow is not
Tuesday.¬p→ ¬q
Contrapositive: If tomorrow is not Tuesday, then today is not
Monday.¬q→ ¬p
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6. WalleTilahun(MSc)walle T
Tautology
Show that ((p→q)∧(q→r))→(p→r) is atautology with out
truth table.
≡((¬p∨q)∧(¬q∨r))→(¬p∨r)
≡ ¬((¬p∨q)∧(¬q∨r))∨(¬p∨r)
≡ ¬((¬p∨q)∧(¬q∨r))∨(¬p∨r)
≡(¬(¬p∧q)∨(¬¬q∧ ¬r))∨(¬p∨r)
≡((p∧ ¬q)∨(q∧ ¬r))∨(¬p∨r)
≡((p∧ ¬q)∨ ¬p)∨((q∧ ¬r)∨r)
≡(p∨ ¬p)∧(¬q∨ ¬p)∨(q∨r)∧(¬r∨r)
≡T∧(¬q∨ ¬p)∨(q∨r)∧T
≡((¬q∨ ¬p)∨q)∨r≡((¬q∨q)∨ ¬p)∨r≡(T∨ ¬p)∨r
≡T∨r=T
∴tautology
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7. WalleTilahun(MSc)walle T
Suppose the uineversal set be given,U=R
Existential Quantifier (∃) and Universal Quantifier (∀)
Example:∃x∀y(x>y)
Interpretation: There exists atleast onexsuch that for ally,xis
greater thany.
Truth Value: False (There is no singlexthat is greater than ally)
Universal Quantifier (∀) and Existential Quantifier (∃)
Example:∀x∃y(x+y= 0)Interpretation: For allx=m∈domain ofx,
there exists aysuch thatm+y= 0.Truth Value: True (e.g.,x=m
andy= 0−m∈Rsatisfy the statement)
Existential Quantifier (∃) and Existential Quantifier (∃)
Example:∃x∃y(x
2
+y
2
= 25)
Interpretation: There existxandysuch thatx
2
+y
2
= 25.
Truth Value: True (e.g.,x= 3 andy= 4 satisfy the statement)
Universal Quantifier (∀) and Universal Quantifier (∀)
Example:∀x∀y(x<y)
Interpretation: For allxand for ally,xis less thany.
Truth Value: False (This statement is not true for all possible values of
x and y)
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8. WalleTilahun(MSc)walle T
Class Work
Question
SupposeU={1,2,3}be the uineversal set,then which one of the
following is not True?
A .∃x∀y(x
2
+ 2y<10)
B.∃x∃y(x
2
+ 2y<10)
C.∃x∃y(x
2
+y
2
≤2xy)
D.∀x∃y(x
2
+ 2y<10)
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9. WalleTilahun(MSc)walle T
Argument Formalization and Evaluation
Given Argument:
1
The butler and the cook are not both innocent.
2
Either the butler is lying or the cook is
3
Therefore, the butler is either lying or guilty.
Formalization:
Let:
p: The butler is innocent
q: The cook is innocent
r: The butler is lying
s: The butler is guilty
Premises:
1
¬(p∧q)
2
(r∨ ¬q)
Conclusion:
1
(r∨s)
using Truth Tables:
using Rule of Inference (EXERCISE!:
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10. WalleTilahun(MSc)walle T
Argument Formalization and Evaluation
Rowpqrs¬(p∧q)r∨¬qr∨s
1TTTT F T T
2TTTF F T T
3TTFT F F T
4TTFF F F F
5TFTT T T T
6TFTF T T T
7TFFT T T T
8TFFF T T F
9FTTT T T T
10FTTF T T F
11FTFT T F T
12FTFF T F T
13FFTT T T T
14FFTF T T T
15FFFT T T T
16FFFF T T F
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11. WalleTilahun(MSc)walle T
Look the premise value in row 8
th
,10
th
,and 16
th
has truth value
simultanuously ,but the conclusion has false value in this row.
∴It is invalid argument.
Definition
LetXbe a set. Thepower setofX, denoted byP(X) or 2
X
is the set
whose elements are all the possible subsets ofX. That is to say,
P(X) ={A|A⊆X}.
SupposeX={1,2,3,4}, then the power set of setAgiven by
P(X) =
{∅,{1},{2},{3},{4},{1,2},{1,3},{1,4},{2,3},{2,4},{3,4}
,{1,2,3},{1,2,4},{1,3,4},{2,3,4},{1,2,3,4}}.
SupposeX={2,{2,4},3}, then list all subset of setA=. . .
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12. WalleTilahun(MSc)walle T
Set Operations
Union (∪):The union of two setsAandBis the set of all elements
that are inA, inB, or in both. The formula for the union is given by:
A∪B={x|x∈Aorx∈B}
Intersection (∩):The intersection of two setsAandBis the set of
all elements that are in bothAandB. The formula for the
intersection is given by:
A∩B={x|x∈Aandx∈B}
Set Difference (−):The set difference of two setsAandBis the
set of all elements that are inAbut not inB. The formula for the set
difference is given by:A−B={x|x∈Aandx/∈B}
Complement (A
′
orA):The complement of a setAwith respect to
the universal setUis the set of all elements inUthat are not inA.
The formula for the complement is given by:
A
′
={x|x∈Uandx/∈A}
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13. WalleTilahun(MSc)walle T
Set Operation Examples
Example 1:LetA={1,2,3,4}andB={3,4,5,6}. FindA∪B,A∩B,
andA−B.
Solution:
A∪B={1,2,3,4,5,6}
A∩B={3,4}
A−B={1,2}
A∆B={1,2,5,6}
Example 2:Let the universal set beU={1,2,3,4,5,6,7,8,9,10}and
A={2,4,6,8}. Find the complement ofA.
Solution:The complement ofAwith respect toUis
A
′
={1,3,5,7,9,10}.
. In a class of 60 students, 25 Students play cricket and 20 students play
tennis, and 10 students play both games. Then, the number of students
who play neither is=. . .
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14. WalleTilahun(MSc)walle T
Venn diagrams with set operations
LetU={1,2, . . . ,12},A={n∈N|n<8},B={n∈N|n<6}, and
C={n∈N|3≤n≤9}. Then we have:
1,23,4910,11,12ABC8576
Some set operations are:
A∪B={n∈N|n<8}
A∩C={n∈N|3≤n<8}
A\B={n∈N|6≤n<8}
C
c
={n∈N|n<3 orn>9}
A∆B= (B\A)∪(A\B)=
A∆(B\A)=
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15. WalleTilahun(MSc)walle T
Definition
LetA,Bbe sets. Therelative complementofBinA, denoted byA\Bis
the set defined by
A\B={x|x∈Aandx/∈B}.
Theorem (De Morgan’s Laws for Sets)
LetA,Bbe sets in the universeΩ. Then
(A∪B)
c
=A
c
∩B
c
, and
(A∩B)
c
=A
c
∪B
c
.
Theorem
LetA,B,Cbe sets. Then
A∩(B∪C) = (A∩B)∪(A∩C), and
A∪(B∩C) = (A∪B)∩(A∪C).
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16. WalleTilahun(MSc)walle T
Proof of Set Identity:A∩(B∪C) = (A∩B)∪(A∩C)
LetA,B, andCbe non-empty sets.
Proof:
Step 1: Consider an elementxinA∩(B∪C).
x∈A∩(B∪C)
This means thatxis in bothAand in the union ofBandC.
Step 2: Break down the expressionx∈A∩(B∪C).
x∈Aandx∈(B∪C)
This implies thatxis inAand in eitherBorC.
Step 3: Consider two cases:
Case 1:xis inB.
x∈Aandx∈B
This implies thatxis in bothAandB, i.e.,x∈A∩B.
Case 2:xis inC.
x∈Aandx∈C
This implies thatxis in bothAandC, i.e.,x∈A∩C.
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17. WalleTilahun(MSc)walle T
Proof CONT’D...A∩(B∪C) = (A∩B)∪(A∩C)
In both cases, we have shown thatxis in eitherA∩BorA∩C, which
proves that
A∩(B∪C)⊆(A∩B)∪(A∩C)
Step 4: Now, consider an elementyinA∩(B∪C). =⇒yis in
eitherA∩Bor inA∩C.
Step 5: Break down the expressiony∈(A∩B)∪(A∩C). We have
two cases to consider:
Case 1:yis inA∩B. =⇒yis in bothAandB, i.e.,yis in the
intersection ofAand the union ofBandC.
Case 2:yis inA∩C. =⇒yis in bothAandC, i.e.,y∈A∩CIn
both cases, we have shown that any element in (A∩B)∪(A∩C) is
also inA∩(B∪C), which proves that
(A∩B)∪(A∩C)⊆A∩(B∪C)
This completes the proof that
A∩(B∪C) = (A∩B)∪(A∩C)
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18. WalleTilahun(MSc)walle T
Proof of Set Identity:A∪(B∩C) = (A∪B)∩(A∪C)
LetA,B, andCbe non-empty sets.
Proof:
Step 1: Consider an elementxinA∪(B∩C).
x∈A∪(B∩C)
This means thatxis in eitherAor in the intersection ofBandC.
Step 2: Break down the expressionx∈A∪(B∩C).
x∈Aorx∈(B∩C)
This implies thatxis inAor in bothBandC.
Step 3: Consider two cases:
Case 1:xis inA. This implies thatxis inA, i.e.,
x∈A∪Bandx∈A∪C
This means thatxis in the union ofAandB, and also in the union of
AandC, i.e.,
x∈(A∪B)∩(A∪C)
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19. WalleTilahun(MSc)walle T
Case 2:xis in bothBandC. This implies thatxis in the
intersection ofBandC, i.e.,
x∈B∩C
In this case, we have shown that any element inA∪(B∩C) is also in
(A∪B)∩(A∪C), which proves that
A∪(B∩C)⊆(A∪B)∩(A∪C)
In both cases, we have shown that any element inA∪(B∩C) is also
in (A∪B)∩(A∪C), which proves that
A∪(B∩C)⊆(A∪B)∩(A∪C)
This completes the proof that
A∪(B∩C) = (A∪B)∩(A∪C)
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20. WalleTilahun(MSc)walle T
Logic
Question 1
Which of the following is a tautology?
A)p∧ ¬p
B)p∨ ¬p
C)p→ ¬p
D)p↔ ¬p
Answer:B)p∨ ¬p
Question 2
Which of the following represents the logical equivalence of (p∧q)∨r?
A)p∧(q∨r)
B)(p∧q)∨(p∧r)
C) (p∨r)∧(q∨r)
D) (p∨q)∧(p∨r)
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21. WalleTilahun(MSc)walle T
Question 1
IfA={1,2,3}, what is the power set ofA?
A){∅,1,2,3,{1,2},{1,3},{2,3},{1,2,3}}
B){∅,{1},{2},{3},{1,2},{1,3},{2,3},{1,2,3}}
C){∅,{∅},{1,2,3}}
D){∅,{∅},{1,2,3},{∅,1,2,3}}
E) None of the above
Answer:B){∅,{1},{2},{3},{1,2},{1,3},{2,3},{1,2,3}}
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22. WalleTilahun(MSc)walle T
Logic
Question 1
What is the validity of the following statement?
(P∧Q)∨(¬P∧Q)
A. Always true
B. Always false
C. True for some cases
D. True for all cases except one
Answer:C. True for some cases
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23. WalleTilahun(MSc)walle T
Set Theory
Question 1
What is the symmetric difference of sets A and B, denoted byA△B?
A.A∪B
B.A∩B
C.A−B
D.A∪B−A∩B
Answer:D.A∪B−A∩B
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24. WalleTilahun(MSc)walle T
Mixed Quantifiers
Question 1
Translate the following sentence into logical symbols: ”There exists a real
number x such that for every real number y, x + y = y + x.”
A. (∃x∈R)(∀y∈R)(x+y=y+x)
B. (∀x∈R)(∃y∈R)(x+y=y+x)
C. (∃x∈R)(∃y∈R)(x+y=y+x)
D. (∀x∈R)(∀y∈R)(x+y=y+x)
Answer:A. (∃x∈R)(∀y∈R)(x+y=y+x)
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25. WalleTilahun(MSc)walle T
UNIT TWO: Real Number system
The remaining portion of this chapter is as a Reading Assignent for you!
Principle of Mahematical Induction
Definition
For a given assertion involving a natural numbern,if
step1:- The asserstion is true forn= 1
step2:If it is true forn=k(k≥1),then the asserstion is true for every
natural nambern.
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26. WalleTilahun(MSc)walle T
Mathematical Induction Examples
Example 1:Prove that 1 + 2 + 3 +. . .+n=
n(n+1)
2
for all positive
integersn.
Proof:We will prove this by mathematical induction.
Base Case:Forn= 1, the left-hand side is 1 and the right-hand side is
1(1+1)
2
= 1. So the equation holds forn= 1.
Inductive Step:Assume that the equation holds for some positive integer
k, i.e., 1 + 2 + 3 +. . .+k=
k(k+1)
2
.
Now we need to show that the equation also holds fork+ 1. Adding
(k+ 1) to both sides of the assumed equation gives:
1 + 2 + 3 +. . .+k+ (k+ 1) =
k(k+ 1)
2
+ (k+ 1)
=
k(k+ 1) + 2(k+ 1)
2
=
(k+ 1)(k+ 2)
2
Thus, the equation holds fork+ 1. By mathematical induction, the
equation holds for all positive integersn.
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27. WalleTilahun(MSc)walle T
Proof of Example 2
Example 2:Prove that 1
2
+ 2
2
+ 3
2
+. . .+n
2
=
n(n+1)(2n+1)
6
for all
positive integersn.
Base Case:Forn= 1, the LHS is 1
2
= 1 and RHS is
1(1+1)(2·1+1)
6
= 1.
So the equation holds forn= 1.
Inductive Step:Assume that the equation holds for some positive integer
k, i.e., 1
2
+ 2
2
+ 3
2
+. . .+k
2
=
k(k+1)(2k+1)
6
.
Now we need to show that the equation also holds fork+ 1. Adding
(k+ 1)
2
to both sides of the assumed equation gives:
1
2
+ 2
2
+ 3
2
+. . .+k
2
+ (k+ 1)
2
=
k(k+ 1)(2k+ 1)
6
+ (k+ 1)
2
=
k(k+ 1)(2k+ 1) + 6(k+ 1)
2
6
=
(k+ 1)(k+ 2)(2k+ 3)
6
Thus, the equation holds fork+ 1. By mathematical induction, the
equation holds for all positive integersn.
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28. WalleTilahun(MSc)walle T
Proof of Example 3
Example 3:Prove that 2
n
>n
2
for all positive integersnsuch thatn≥5
by mathematical induction.
Base Case:Forn= 5, we have 2
5
= 32 and 5
2
= 25, so 2
5
>5
2
. The
base case holds.
Inductive Step:Assume that the inequality holds for some positive
integerk≥5, i.e., 2
k
>k
2
.
Now we need to show that the inequality also holds fork+ 1. Multiplying
both sides of the assumed inequality by 2 gives:
2
k+1
= 2·2
k
>2·k
2
Sincek≥5, we have 2k>k
2
, so 2·k
2
>k
2
. Therefore, 2
k+1
>k
2
. By
mathematical induction, the inequality holds for all positive integersnsuch
thatn≥5.
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29. WalleTilahun(MSc)walle T
Complex Numbers
Introduction to Complex Numbers
A complex number is a number that can be expressed in the forma+bi,
whereaandbare real numbers, andiis the imaginary unit (with the
propertyi
2
=−1).
Definition of Complex Numbers
A complex numberzcan be written asz=a+bi, whereais the real part
andbis the imaginary part ofz.
Operations with Complex Numbers
Addition: (a+bi) + (c+di) = (a+c) + (b+d)i
Subtraction: (a+bi)−(c+di) = (a−c) + (b−d)i
Multiplication: (a+bi)×(c+di) = (ac−bd) + (ad+bc)i
Division:
a+bi
c+di
=
(ac+bd)+(bc−ad)i
c
2
+d
2
Set By:Walle Tilahun(MSc) Email: [email protected] If there is error reply me!Math 1011 teaching material for section IProofs and Examples are not included under this Material please try to refer your module!January 22, 2024

30. WalleTilahun(MSc)walle T
Proof of Commutative Property of Addition and
Subtraction for Complex Numbers
Letz1=a1+b1iandz2=a2+b2ibe complex numbers.
Addition:
Step 1: Consider the sumz1+z2.
z1+z2= (a1+b1i) + (a2+b2i)
Step 2: Rearrange the terms in the sum.
z1+z2= (a1+a2) + (b1+b2)i
Step 3: Consider the sumz2+z1.
z2+z1= (a2+b2i) + (a1+b1i)
Step 4: Rearrange the terms in the sum.
z2+z1= (a2+a1) + (b2+b1)i
Step 5: Compare the results from Steps 2 and 4. We have
z1+z2=z2+z1, which proves the commutative property of addition for
complex numbers.
Subtraction:
Step 1: Consider the differencez1−z2.
z1−z2= (a1+b1i)−(a2+b2i)
Step 2: Rearrange the terms in the difference.
z1−z2= (a1−a2) + (b1−b2)i
Step 3: Consider the differencez2−z1.
z2−z1= (a2+b2i)−(a1+b1i)
Step 4: Rearrange the terms in the difference.
z2−z1= (a2−a1) + (b2−b1)i
Step 5: Compare the results from Steps 2 and 4. We have
z1−z2̸=z2−z1, which shows that subtraction is not commutative for
complex numbers.
Set By:Walle Tilahun(MSc) Email: [email protected] If there is error reply me!Math 1011 teaching material for section IProofs and Examples are not included under this Material please try to refer your module!January 22, 2024

31. WalleTilahun(MSc)walle T
Complex Numbers
Conjugate of Complex Numbers
The conjugate of complex numberz=a+biis denoted byz=a−bi.
Triangular Inequality . . . proof!
For two complex numbersz1=a1+b1iandz2=a2+b2i, the triangular
inequality states that:
|z1+z2| ≤ |z1|+|z2|
. Example: Letz1= 3 + 4iandz2= 1−2i. We have:
|z1+z2|=|4 + 2i|=
√
20
|z1|+|z2|=|3 + 4i|+|1−2i|= 5 +
√
5
Since
√
20≤5 +
√
5, the triangular inequality holds
Operations with Complex Numbers
The Additive inverse of:z= (x+yi) is−z= (−x−yi)
The Multiplicative inveverse of:z= (x+yi) is given
by
1
z
=
1
x+yi
=
x−yi
√
x
2
+y
2
Set By:Walle Tilahun(MSc) Email: [email protected] If there is error reply me!Math 1011 teaching material for section IProofs and Examples are not included under this Material please try to refer your module!January 22, 2024

32. WalleTilahun(MSc)walle T
Proof of Triangle Inequality
|z1+z2| ≤ |z1|+|z2|
Letz1=a1+b1iandz2=a2+b2ibe two complex numbers.
|z1+z2|
2
= (a1+a2)
2
+ (b1+b2)
2
=a
2
1+ 2a1a2+a
2
2+b
2
1+ 2b1b2+b
2
2
= (a
2
1+b
2
1) + 2(a1a2+b1b2) + (a
2
2+b
2
2)
=|z1|
2
+ 2Re(z1z2) +|z2|
2
≤ |z1|
2
+ 2|z1||z2|+|z2|
2
= (|z1|+|z2|)
2
Taking the square root of both sides, we get:|z1+z2| ≤ |z1|+|z2|
Therefore, the triangle inequality holds for complex numbers.
Set By:Walle Tilahun(MSc) Email: [email protected] If there is error reply me!Math 1011 teaching material for section IProofs and Examples are not included under this Material please try to refer your module!January 22, 2024

33. WalleTilahun(MSc)walle T
Proof of Inequality|z1−z2| ≥ |z1| − |z2|
To prove:|z1−z2| ≥ |z1| − |z2|
Proof:
|z1−z2|
2
= (z1−z2)(z1−z2)
=|z1|
2
−z1z2−z1z2+|z2|
2
=|z1|
2
−2Re(z1z2) +|z2|
2
=|z1|
2
−2Re(z1z2) +|z2|
2
≥ |z1|
2
−2|z1||z2|+|z2|
2
=|z1|
2
−2|z1||z2|+|z2|
2
= (|z1| − |z2|)
2
Taking the square root of both sides gives:
|z1−z2| ≥ |z1| − |z2|
Adding|z2|to both sides yields the desired inequality:
|z1−z2| ≥ |z1| − |z2|
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34. WalleTilahun(MSc)walle T
Solution: Operations on Complex Numbers
Let’s perform some operations on the complex numbersz1= 2 + 3iand
z2= 1−2i.
Addition:We havez1+z2= (2 + 1) + (3−2)i= 3 +i.
Subtraction:We havez1−z2= (2−1) + (3 + 2)i= 1 + 5i.
Multiplication:We havez1·z2= (2·1−3·2) + (2·1 + 3·1)i=−4 + 5i.
Division:We have
z1
z2
=
(2+3i)
(1−2i)
. To simplify this, we multiply the
numerator and denominator by the conjugate of the denominator.
Exrecise 1:3i
87
+i
125
−i
7
+ (5i
42
−7i
37
) =. . .
Exrecise 2:Findz∈Csuch thatRe(z(1 +i)) +zz= 0
Set By:Walle Tilahun(MSc) Email: [email protected] If there is error reply me!Math 1011 teaching material for section IProofs and Examples are not included under this Material please try to refer your module!January 22, 2024

35. WalleTilahun(MSc)walle T
Solution: Modulus and Argument
Let’s find the modulus and argument of the complex numberz= 3 + 4i.
Modulus:We have|z|=
√
3
2
+ 4
2
= 5.
Argument:We have arg(z) = tan
−1
Γ
4
3
∆
.
Therefore, the modulus ofzis 5 and the argument ofzis tan
−1
Γ
4
3
∆
.
|z|=
√
zz
Suppose the equation of the cirecle is given by|z−3 + 2i|= 4,then
find the center of circle and radius of the circle.
solution
|z−3 + 2i|= 4 =⇒ |z−(3−2i)|= 4.
This represents a circle with center (3,−2) and radius 4.
Set By:Walle Tilahun(MSc) Email: [email protected] If there is error reply me!Math 1011 teaching material for section IProofs and Examples are not included under this Material please try to refer your module!January 22, 2024

36. WalleTilahun(MSc)walle T
Complex Plane and Modulus/Argument
ℜℑz=a+biab|z|θ
Modulus:|z|=
√
a
2
+b
2
Argument:θ= arctan
Γ
b
a
∆
Set By:Walle Tilahun(MSc) Email: [email protected] If there is error reply me!Math 1011 teaching material for section IProofs and Examples are not included under this Material please try to refer your module!January 22, 2024

37. WalleTilahun(MSc)walle T
Polar Form of Complex Numbers
Modulus and Argument
The modulus of a complex numberz=a+biis given by
r=|z|=
√
a
2
+b
2
, and the argument ofzis given byθ= arctan
Γ
b
a
∆
.
Polar Form
A complex numberz=a+bican also be represented in polar form as
z=r(cosθ+isinθ), whereris the modulus andθis the principal
argument ofz.
z=re
iθ
is called Eulers formula.
De Moivre’s Theorem
De Moivre’s Theorem states that for any complex number
z=r(cosθ+isinθ), and any positive integern, we have
z
n
=r
n
(cosnθ+isinnθ).
Set By:Walle Tilahun(MSc) Email: [email protected] If there is error reply me!Math 1011 teaching material for section IProofs and Examples are not included under this Material please try to refer your module!January 22, 2024

38. WalleTilahun(MSc)walle T
Extraction of Roots and Triangular Inequality
Extraction of Roots
Thenth roots of a complex numberz=r(cosθ+isinθ) can be found
using the formula:
zk=
n
√
r
`
cos
`
θ+ 2kπ
n
´
+isin
`
θ+ 2kπ
n
´´
or
Zk=r
1
ne
i(
θ+2kπ
n
)
wherek= 0,1,2, ...,n−1.
For two complex numbersz1= 1 +
√
2iandz2= 1 + 2i, then evaluate
a. arg(z1+z2) = b.Im(z1+z2)= c.Im(z2z2)= d.z2+z2=
Set By:Walle Tilahun(MSc) Email: [email protected] If there is error reply me!Math 1011 teaching material for section IProofs and Examples are not included under this Material please try to refer your module!January 22, 2024

39. WalleTilahun(MSc)walle T
Solution: Extraction ofn
th
Root of Complex Number
Example 1:Let’s find the fourth roots of the complex number
z= 16
Γ
cos
π
3
+isin
π
3
∆
.
Solution:The modulus ofzis|z|=
√
16
2
= 16. The argument ofzis
π
3
.
The fourth roots ofzare given by:
wk=
4
√
16
`
cos
`
π
3
+ 2kπ
4
´
+isin
`
π
3
+ 2kπ
4
´´
fork= 0,1,2,3.
Plugging in the values ofk, we get:
w0= 2
ı
cos
π
12
+isin
π
12
ȷ
w1= 2
`
cos
7π
12
+isin
7π
12
´
w2= 2
`
cos
13π
12
+isin
13π
12
´
w3= 2
`
cos
19π
12
+isin
19π
12
´
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40. WalleTilahun(MSc)walle T
Solution: Finding Center and Radius of Circle for Complex
Number
Example 2:Given the complex numberz=−3 + 4i, find the center and
radius of the circle in the complex plane.
Solution:The real part ofzis−3 and the imaginary part is 4. So, the
center of the circle is at the point (−3,4).
The modulus ofzis|z|=
p
(−3)
2
+ 4
2
= 5. Therefore, the radius of the
circle is 5.
Hence, the center of the circle for the complex numberz=−3 + 4iis at
(−3,4) and the radius is 5.
Exrecise 1:find the cube root ofz=−8
Exrecise :find the square root ofz=−1−i
Exrecise :Ifz= (1 +i)
6
,write in polar form.
Set By:Walle Tilahun(MSc) Email: [email protected] If there is error reply me!Math 1011 teaching material for section IProofs and Examples are not included under this Material please try to refer your module!January 22, 2024

41. WalleTilahun(MSc)walle T
Complex Numbers Quiz
Question 1
What is the conjugate of the complex numberz= 3 + 4i?
A) 3−4i
B)−3 + 4i
C)−3−4i
D) 3 + 4i
Answer:A) 3−4i
Question 2
What is the modulus of the complex numberz= 3 + 4i?
A) 7
B) 5
C)
√
7
D)
√
25
Answer:B) 5
Set By:Walle Tilahun(MSc) Email: [email protected] If there is error reply me!Math 1011 teaching material for section IProofs and Examples are not included under this Material please try to refer your module!January 22, 2024

42. WalleTilahun(MSc)walle T
Complex Numbers Quiz
What is the result of dividing the complex numberz1= 3 + 4iby
z2= 1−i?
A)−1 + 2i
B)−1−i
C) 2 +i
D) 2−i
Answer:D) 2−i
What is the real part of the complex numberz=
2i
3+4i
?
A) 3/5
B) 8/5
C)−8/5
D) 0
Answer:A) 8/5
Set By:Walle Tilahun(MSc) Email: [email protected] If there is error reply me!Math 1011 teaching material for section IProofs and Examples are not included under this Material please try to refer your module!January 22, 2024

43. WalleTilahun(MSc)walle T
Complex Numbers Quiz
Question 5
What is the argument (in radians) of the complex numberz=−1 +i
using De Moivre’s formula?
A)
π
4
B)
π
2
C)
π
6
D)
π
3
Answer:A)
π
4
The Multiplicative inverse of 3 + 4iis. . . .
A.
−3−4i
−3+4i
B.
3−4i
5
C.
−3−4i
5
D.
−3+4i
5
Answer:B.
3−4i
5
Set By:Walle Tilahun(MSc) Email: [email protected] If there is error reply me!Math 1011 teaching material for section IProofs and Examples are not included under this Material please try to refer your module!January 22, 2024

44. WalleTilahun(MSc)walle T
Multiple Choice Question
What are the fourth roots of the complex numberz=−32?
A) 2 + 2i
B)−2 + 2i
C)−2−2i
D) 2−2i
E) All of the above
Answer:E) All of the above
Explanation:The principal fourth root ofz=−32 is 2 + 2i. The other
three fourth roots can be obtained by rotating the principal root by
π
2
,π,
and
3π
2
, resulting in−2 + 2i,−2−2i, and 2−2i. Therefore, all of the
given options are correct.
Set By:Walle Tilahun(MSc) Email: [email protected] If there is error reply me!Math 1011 teaching material for section IProofs and Examples are not included under this Material please try to refer your module!January 22, 2024

45. WalleTilahun(MSc)walle T
Complex Numbers
Convert the complex number 3 + 4ifrom rectangular form to polar form.
A. 5e
i
π
3
B. 5e
i
π
4
C. 5e
i
π
6
D. 5e
i
π
2
Answer:B. 5e
i
π
4
The Additive inverse of 3 + 4iis. . . .
A.−3 + 4i
B.−3−4i
C. 3i−4
D. 3−4i
Answer:B.−3−4i
Set By:Walle Tilahun(MSc) Email: [email protected] If there is error reply me!Math 1011 teaching material for section IProofs and Examples are not included under this Material please try to refer your module!January 22, 2024

46. WalleTilahun(MSc)walle T
self exercise
Example 2:Given the complex numberz1= 1 +iandz2=−
√
3−i,
then find the principal argument given below.
A.Arg(z1z2)=. . .
B.Arg(z1) +Arg(z2)=. . .
C.Arg(
z1
z2
)=. . .
D.Arg(z1)−Arg(z2)=. . .
NB:-Argz∈(−π, π),
Argz is called principal Argument ofz.
Solution:
Set By:Walle Tilahun(MSc) Email: [email protected] If there is error reply me!Math 1011 teaching material for section IProofs and Examples are not included under this Material please try to refer your module!January 22, 2024

47. WalleTilahun(MSc)walle T
self exercise
1. find the argument of the following complex numbers:
Z=
i
−1−i
Z= (
√
3−i)
4
2.show that|e
iθ
|= 1
3. solve the following equations
z
3
2= 8i
z
2
+ 4i= 0
4. write by the rectangular form and using Euler’s formula
ifz= 2
60
(cos10π+sin10π)
5.sketch the following set of points determined by conditions given
below|z−1 +i|= 4
6. Write the modulus,argument,real,and imaginary part
ofi
11
+ 2i
5
−i
7
+i(5i
2
−7i
3
+ 3)
Set By:Walle Tilahun(MSc) Email: [email protected] If there is error reply me!Math 1011 teaching material for section IProofs and Examples are not included under this Material please try to refer your module!January 22, 2024