6.6 DeMoivre’s Theorem

I. Trigonometric Form of Complex Numbers

A.) The standard form of the complex number

is very similar to the component form of a vector

If we look at the trigonometric form of v, we can see

.ai bj v

z a b i

cos sini j v v

P (a, b)

B.) If we graph the complex z = a + bi on the complex plane, we can see the similarities with the polar plane.

z = a + bi

θ

r

a

b

C.) If we let and then,

where

sin b r cosa r

2 2 ,r z a b

cos sinz a b r r i i

tan and 1b

a i

D.) Def. – The trigonometric form of a complex number z is given by

Where r is the MODULUS of z and θ is the ARGUMENT of z.

cos sin

or

z r

z rcis

i

E.) Ex.1 - Find the trig form of the following:1 3

2.) 2 2

i1.) 2i

2 2

1

0 2 2

2tan

0 2

0 2

(cos sin )

2 cos sin2 2

22

r

z i

r

cis

i

i

22

1

1 31

2 2

5tan 3

3

1 3

2 2(cos sin )

5 51 cos sin

3 3

5

3

r

z i

r

cis

i

i

A.) Let .

Mult.-

Div. -

1 1 1 1 2 2 2 2cos sin and cos sinz r i z r i

1 2 1 2 1 2 1 2cos sinz z r r i

II. Products and Quotients

1 11 2 1 2

2 2

cos sinz r

iz r

DERIVE THESE!!!!

B.) Ex. 2 – Given .

find 11 2

2

and z

z zz

1 2 5 2 cos 120 sin 120

1 310

2 2

5 5 3

z z i

i

i

1

2

5cos 90 sin 90

2

50

25

2

zi

z

i

i

1

2

5 cos15 sin15 &

2 cos105 sin105

z i

z i

III. Powers of Complex Numbers

A.) DeMoivre’s (di-’moi-vərz) Theorem –

If z = r(cosθ + i sinθ) and n is a positive integer, then,

(cos sin ) cos sinnn nz r i r n i n

Why??? – Let’s look at z2-

2z z z

2 cos sin cos sinz r i r i

2 2 cos sin cos sinz r i i

2 2 2 2 2cos 2 cos sin sinz r i i

2 2 cos 2 sin 2z r i

2 2 2 2cos 1 sin 2cos sinz r i

B.) Ex. 3 – Find by “Foiling” 3

1 3i

1 3 1 3 1 3i i i

1 2 3 3 1 3i i

2 2 3 1 3i i

8

2 6

C.) Ex. 4– Now find using DeMoivre’s Theorem

3

1 3i

2 3

r

3

3 2 cos sin3 3

z i

3 32 cos3 sin 33 3

z i

3 8( 1 0) 8z

3 8 cos sinz i

D.) Ex. 5 –Use DeMoivre’s Theorem to simplify10

3 3

2 2i

2

3 62 =

2 2 4r

10

10 6 10 10cos sin

2 4 4z i

510

10

6 5 5cos sin

2 2 2z i

510

10

6

2z i

IV. nth Roots of Complex Numbers

A.) Roots of Complex Numbers –

v = a + bi is an nth root of z iff vn = z .

If z = 1, then v is an nth ROOT OF UNITY.

B.) If , then the n distinct complex numbers

cos sinz r i

+ 2 + 2cos sinn k k

rn n

i

Where k = 0, 1, 2, …, n-1 are the nth roots of the complex number z.

C.) Ex. 6- Find the 4th roots of

41

2164

z cis

16 .2

z cis

4

3

4216

4z cis

42

2216

4z cis

44

6216

4z cis

1 28

z cis

3

92

8z cis

2

52

8z cis

4

132

8z cis

A.) Ex. 7 - Find the cube roots of -1.

2( 1) 1 0z z z

3 1z

V. Finding Cube Roots

3 1 0z 1 1 4(1)(1)1 or

2(1)z z

1 3 1 31, ,

2 2 2 2z i i

Now....

1

1 3cos sin

3 3 2 2z i

i

1 0z i

1 cos sinz i

2

3 3cos sin 1 0

3 3z i i

3

5 5 1 3cos sin

3 3 2 2z i

i

Plot these points on the complex plane. What do you notice about them?

1

-1

2

1z

2z

3z

Equidistant from the origin and equally spaced about the origin.

VI. Roots of Unity

A.) Any complex root of the number 1 is also known as a ROOT OF UNITY.

B.) Ex. 8 - Find the 6 roots of unity.

1 0 0z i cis

1

0

6z cis

2

0 2

6z cis

3

0 4

6z cis

1

3cis

1 3

2 2i

2

3cis

1 3

2 2i

4

0 6

6z cis

6

0 10

6z cis

5

0 8

6z cis

cis 1

4

3cis

1 3

2 2i

5

3cis

1 3

2 2i