a6 - complex numbers joseph.pdf

The Set of Complex Numbers

Mathematics 17

Institute of Mathematics, University of the Philippines-Diliman

Lecture 6

Math 17 (UP-IMath) The Set of Complex Numbers Lec 6 1 / 15

Outline

1 The Set of Complex NumbersProperties of Complex NumbersConjugate of a Complex NumberOperations Involving Complex Numbers

Addition and Subtraction of Complex NumbersMultiplication and Division of Complex Numbers

The Powers of i



Recall: There is no real number b such that b2 = −1.

We define i to be a number such that i2 = −1.

A complex number is a number of the form a + b · i, where a and b arereal numbers.

Definition

The set of complex numbers C is defined as:

C = {a + bi | a, b ∈ R, i2 = −1}



Properties of Complex Numbers

Let z = a + bi ∈ C.

Standard or Rectangular Form: z = a + bi

Real Part of z: Re(z) = a

Imaginary Part of z: Im(z) = b

Example:z = 3− 7i

Re(z) = 3

Im(z) = −7



Properties of Complex Numbers

Given z = a + bi, z ∈ Cif b = 0, z is a real number: a = a + 0i.

if b 6= 0, z is an imaginary number.

if a = 0 and b 6= 0, z is a pure imaginary number: bi = 0 + bi.

Examples:

5 is a real number.

2 + 3i is an imaginary number, but is not a pure imaginary number.

7i is a pure imaginary number.


Principal Square Root of −p

Definition

If p > 0, then the principal square root of −p, denoted by√−p, is given by

√−p = i

√p

Examples:√−4 = i

√4 = 2i

3i and −3i are square roots of -9 since (−3i)2 = −9 and (3i)2 = −9.The principal square root of −9 is 3i. That is,

√−9 = 3i.


Conjugate of a Complex Number

Definition

The conjugate of z = a + bi:

z̄ = a + bi = a− bi.

Examples:

7 + 2i

= 7− 2i32 −

12 i = 3

2 + 12 i

9 = 9

5i = −5i

i− 1 = −1− i



Definition


z̄ = a + bi = a− bi.

Examples:

7 + 2i = 7− 2i

32 −

12 i = 3

2 + 12 i

9 = 9

5i = −5i

i− 1 = −1− i



Definition


z̄ = a + bi = a− bi.

Examples:

7 + 2i = 7− 2i32 −

12 i

= 32 + 1

2 i

9 = 9

5i = −5i

i− 1 = −1− i



Definition


z̄ = a + bi = a− bi.

Examples:

7 + 2i = 7− 2i32 −

12 i = 3

2 + 12 i

9 = 9

5i = −5i

i− 1 = −1− i



Definition


z̄ = a + bi = a− bi.

Examples:

7 + 2i = 7− 2i32 −

12 i = 3

2 + 12 i

9

= 9

5i = −5i

i− 1 = −1− i



Definition


z̄ = a + bi = a− bi.

Examples:

7 + 2i = 7− 2i32 −

12 i = 3

2 + 12 i

9 = 9

5i = −5i

i− 1 = −1− i



Definition


z̄ = a + bi = a− bi.

Examples:

7 + 2i = 7− 2i32 −

12 i = 3

2 + 12 i

9 = 9

5i

= −5i

i− 1 = −1− i



Definition


z̄ = a + bi = a− bi.

Examples:

7 + 2i = 7− 2i32 −

12 i = 3

2 + 12 i

9 = 9

5i = −5i

i− 1 = −1− i



Definition


z̄ = a + bi = a− bi.

Examples:

7 + 2i = 7− 2i32 −

12 i = 3

2 + 12 i

9 = 9

5i = −5i

i− 1

= −1− i



Definition


z̄ = a + bi = a− bi.

Examples:

7 + 2i = 7− 2i32 −

12 i = 3

2 + 12 i

9 = 9

5i = −5i

i− 1 = −1− i


Operations on Complex Numbers

1. Two complex numbers a + bi and c + di are equal if a = c and b = d.

2. The axioms for the real number system hold for C, except for theorder axioms.

3. There is no concept of ‘order’ in C.

4. Identity Element of Addition in C: 0 = 0 + 0i

5. Identity Element of Multiplication in C: 1 = 1 + 0i


Addition and Subtraction of Complex Numbers

Definition

Let z1 = a + bi, z2 = c + di ∈ C.Then

z1 + z2 = (a + bi) + (c + di)

= (a + c) + (b + d)i

z1 − z2 = (a + bi)− (c + di)

= (a− c) + (b− d)i



Example: Let z1 = 3 + 2i and z2 = 4− 3i.

1. z1 + z2

z1 + z2 = (3 + 2i) + (4 +−3i)= (3 + 4) + (2 + (−3))i= 7− i

2. z1 − z2z1 − z2 = (3 + 2i)− (4 + 3i)

= (3− 4) + (2− (−3))i= −1 + 5i




1. z1 + z2z1 + z2 = (3 + 2i) + (4 +−3i)

= (3 + 4) + (2 + (−3))i= 7− i

2. z1 − z2z1 − z2 = (3 + 2i)− (4 + 3i)

= (3− 4) + (2− (−3))i= −1 + 5i




1. z1 + z2z1 + z2 = (3 + 2i) + (4 +−3i)

= (3 + 4) + (2 + (−3))i

= 7− i

2. z1 − z2z1 − z2 = (3 + 2i)− (4 + 3i)

= (3− 4) + (2− (−3))i= −1 + 5i




1. z1 + z2z1 + z2 = (3 + 2i) + (4 +−3i)

= (3 + 4) + (2 + (−3))i= 7− i

2. z1 − z2z1 − z2 = (3 + 2i)− (4 + 3i)

= (3− 4) + (2− (−3))i= −1 + 5i




1. z1 + z2z1 + z2 = (3 + 2i) + (4 +−3i)

= (3 + 4) + (2 + (−3))i= 7− i

2. z1 − z2

z1 − z2 = (3 + 2i)− (4 + 3i)= (3− 4) + (2− (−3))i= −1 + 5i




1. z1 + z2z1 + z2 = (3 + 2i) + (4 +−3i)

= (3 + 4) + (2 + (−3))i= 7− i

2. z1 − z2z1 − z2 = (3 + 2i)− (4 + 3i)

= (3− 4) + (2− (−3))i= −1 + 5i




1. z1 + z2z1 + z2 = (3 + 2i) + (4 +−3i)

= (3 + 4) + (2 + (−3))i= 7− i

2. z1 − z2z1 − z2 = (3 + 2i)− (4 + 3i)

= (3− 4) + (2− (−3))i

= −1 + 5i




1. z1 + z2z1 + z2 = (3 + 2i) + (4 +−3i)

= (3 + 4) + (2 + (−3))i= 7− i

2. z1 − z2z1 − z2 = (3 + 2i)− (4 + 3i)

= (3− 4) + (2− (−3))i= −1 + 5i


Multiplication and Division of Complex Numbers

Definition

If z1 = a + bi, z2 = c + di ∈ C, then

z1 · z2 = (a + bi) · (c + di)

= (ac− bd) + (ad + bc)i

z1z2

=a + bi

c + di· c− di

c− di

=ac + bd

c2 + d2+

bc− ad

c2 + d2i, z2 6= 0

Note: The laws of integer exponents apply to complex numbers.



Example: Let z1 = 2− 3i and z2 = 1 + 5i.

1. z1 · z2

z1 · z2 = (2 + (−3)i) · (1 + 5i)= (2 · 1− (−3) · 5) + (2 · 5 + (−3) · 1)i= (2− (−15)) + (10− 3)i= 17− 7i

2.z1z2

z1z2

=2− 3i

1 + 5i· 1− 5i

1− 5i

=2− 15 + (−3− 10)i

1 + 25

=−13− 13i

26= −1

2− 1

2i




1. z1 · z2

z1 · z2 = (2 + (−3)i) · (1 + 5i)

= (2 · 1− (−3) · 5) + (2 · 5 + (−3) · 1)i= (2− (−15)) + (10− 3)i= 17− 7i

2.z1z2

z1z2

=2− 3i

1 + 5i· 1− 5i

1− 5i

=2− 15 + (−3− 10)i

1 + 25

=−13− 13i

26= −1

2− 1

2i




1. z1 · z2

z1 · z2 = (2 + (−3)i) · (1 + 5i)= (2 · 1− (−3) · 5) + (2 · 5 + (−3) · 1)i

= (2− (−15)) + (10− 3)i= 17− 7i

2.z1z2

z1z2

=2− 3i

1 + 5i· 1− 5i

1− 5i

=2− 15 + (−3− 10)i

1 + 25

=−13− 13i

26= −1

2− 1

2i




1. z1 · z2

z1 · z2 = (2 + (−3)i) · (1 + 5i)= (2 · 1− (−3) · 5) + (2 · 5 + (−3) · 1)i= (2− (−15)) + (10− 3)i

= 17− 7i

2.z1z2

z1z2

=2− 3i

1 + 5i· 1− 5i

1− 5i

=2− 15 + (−3− 10)i

1 + 25

=−13− 13i

26= −1

2− 1

2i




1. z1 · z2

z1 · z2 = (2 + (−3)i) · (1 + 5i)= (2 · 1− (−3) · 5) + (2 · 5 + (−3) · 1)i= (2− (−15)) + (10− 3)i= 17− 7i

2.z1z2

z1z2

=2− 3i

1 + 5i· 1− 5i

1− 5i

=2− 15 + (−3− 10)i

1 + 25

=−13− 13i

26= −1

2− 1

2i




1. z1 · z2

z1 · z2 = (2 + (−3)i) · (1 + 5i)= (2 · 1− (−3) · 5) + (2 · 5 + (−3) · 1)i= (2− (−15)) + (10− 3)i= 17− 7i

2.z1z2

z1z2

=2− 3i

1 + 5i

· 1− 5i

1− 5i

=2− 15 + (−3− 10)i

1 + 25

=−13− 13i

26= −1

2− 1

2i




1. z1 · z2

z1 · z2 = (2 + (−3)i) · (1 + 5i)= (2 · 1− (−3) · 5) + (2 · 5 + (−3) · 1)i= (2− (−15)) + (10− 3)i= 17− 7i

2.z1z2

z1z2

=2− 3i

1 + 5i· 1− 5i

1− 5i

=2− 15 + (−3− 10)i

1 + 25

=−13− 13i

26= −1

2− 1

2i


The Powers of i

Powers of i

Let n ∈ N.

If r is the remainder when n is divided by 4, then

in

=

1 if r = 0i if r = 1−1 if r = 2−i if r = 3


The Powers of i

Powers of i

Let n ∈ N.

If r is the remainder when n is divided by 4, then

in =

1 if r = 0i if r = 1−1 if r = 2−i if r = 3


The Powers of i

Example:

−7i13 + 10i6 − 5i3 + 4i

= −7i4·3+1 + 10i4·1+2 − 5i4·0+3 + 4i

= −7i− 10 + 5i + 4i

= −10 + 2i


The Powers of i

Example:

−7i13 + 10i6 − 5i3 + 4i = −7i4·3+1 + 10i4·1+2 − 5i4·0+3 + 4i

= −7i− 10 + 5i + 4i

= −10 + 2i


Exercise:

Perform the following operations and simplify.

1 3i(i2012 − i67 + 5i5 − i−2)

2

(−1

2+

√3

2i

)(√2

2−√

2

2i

)3

3i− 2

3i + 2

47 + i− 4(3− i)

6− 5i51

52− i75 − 2(i + 1)

2−√−4


a6 - complex numbers joseph.pdf

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