9.3Complex Numbers;

The Complex Plane;Polar Form of Complex Numbers

Complex Numbers are numbers in the form of a biwhere a and b are real numbers and i, the imaginary unit, is defined as follows:

2 1i 1i And the powers of i are as follows:

1i i2 1i 3 2 1i i i i i 4 1i

The value of in, where n is any number can be found by dividing n by 4 and then dealing only with the remainder. Why?

Examples:

18i ? 18 4 4 2with a remainder of 18 2i i Then from the chart on the previous slide18 2 1i i

1)

2) 27i ? 27 4 6 3with a remainder of 27 3i i27 3i i i

Then from the chart on the previous slide

In a complex number a bi

a is the real part and bi is the imaginary part.

When b=0, the complex number is a real number.

When a0, and b0, as in 5+8i, the complex number is an imaginary number.

When a=0, and b0, as in 5i, the complex number is a pure imaginary number.

Lesson Overview 9-5A

Lesson Overview 9-5B

5-Minute Check Lesson 9-6A

Real Axis

Imaginary Axis

O

z a bi

The Complex Plane

Let z a bi be a complex number.

The magnitude or modulus of z, denoted by z is defined

As the distance from the origin to the point (x, y).

2 2z a b

Real Axis

Imaginary Axis

O x

y

|z|

z a bi

sinicosr

is sometimes abbreviated as

cisr

4

-3 Real Axis

Imaginary Axis

z =-3 + 4i

z = -3 + 4i is in Quadrant II

x = -3 and y = 4

4

-3

z =-3 + 4i

5r

Find the reference angle () by solving

x

ytan

3

4tan

3

4tan 1

13.53

4

-3

z =-3 + 4i

5r

Find r: 2213 r

2r

4

-3 Real Axis

Imaginary Axis

i3

3

12

Find the reference angle () by solving

x

ytan

3

1tan

30

3

12

3

11 tan

3

12

33030360 sinicosrz

3303302 sinicosz

Find the cosine of 330 and substitute the value.

Find the sine of 330 and substitute the value.

Distribute the r

6

5

6

52

sinicos

Write in standard (rectangular) form.

2

3

6

5

cos

2

1

6

5

sin

i

2

1

2

32 i 3

Lesson Overview 9-7A

Product Theorem

Lesson Overview 9-7B

Quotient Theorem

5-Minute Check Lesson 9-8A

5-Minute Check Lesson 9-8B

Powers and Roots of Complex Numbers

DeMoivre’s Theorem

81 81 3

2 2i

4326 i

What if you wanted to perform the operation below?

Lesson Overview 9-8A

Lesson Overview 9-8B

Theorem Finding Complex Roots

roots

Find the complex fifth roots of

The five complex roots are:

for k = 0, 1, 2, 3,4 .

2 2r a b

32 0i2 232 0 232 32

1 0

32tan 1 0tan 0

32 0 0cos i sin

5 0 360 0 36032

5 5 5 5

k kcos i sin

2 72 72cos k i sin k

0k 2 0 0cos i sin 2 1 0i 2

1k

2 72 72cos k i sin k

2 72 72cos i sin 62 1 90. . i

2k 2 144 144cos i sin 1 62 1 18. . i

3k 2 216 216cos i sin 1 62 1 18. . i

4k 2 288 288cos i sin 62 1 90. . i

The roots of a complex number a cyclical in nature.

That is, when the points are plotted on a polar plane or a complex plane, the points are evenly spaced around the origin

4

2

-2

-4

-5 5

Complex Plane

6

4

2

2

4

5 5 10

Polar plane

6

4

2

2

4

5 5 10

Polar plane

6

4

2

2

4

5 5 10

4

2

-2

-4

-5 5

To find the principle root, use DeMoivre’s theorem using rational exponents.

That is, to find the principle pth root of

a bi Raise it to the power1

p

Example

Find 3 i

First express as a complex number in standard form. i0 i

Then change to polar form 1r 2 2r a b

btan

a

1

0tan 1 1

0tan 90

1 90 90cos i sin

You may assume it is the principle root you are seeking unless specifically stated otherwise.

1 90 90cos i sin Since we are looking for the cube root, use DeMoivre’s Theorem and raise it to the power1

31

3 1 11 90 90

3 3cos i sin

1 30 30cos i sin

3 11

2 2i

3 1

2 2i

Example:

Find the 4th root of 2 i

1

42 i

Change to polar form 5 153 4 153 4cos . i sin .

Apply DeMoivre’s Theorem

1

41 1

5 153 4 153 44 4

cos . i sin .

1 22 38 4 38 4. cos . i sin .

96 76. . i