6.5 Trig. Form of a Complex Number

22 babiaz +=+=

Ex. Z = -2 + 5i ( )5,2−

22 5)2( +−=z

29=z -2

5

Trigonometric Form of a Complex Number

( )θθ sincos irz +=r = the hypotenuse and theta = the angle.

Ex. iz 322−−= Find r (the hyp) & theta

r = 4

===2

32

.

.tan

adj

oppθ

1

3

060=∴θ ref. angle

°=∴ 2403

4or

πθ

)3

4sin

3

4(cos4

ππiz +=

Write in trig. form

2−

32−

Write the complex number in standard form a + bi.

Ex. ⎥⎦

⎤⎢⎣

⎡⎟⎠

⎞⎜⎝

⎛−+⎟⎠

⎞⎜⎝

⎛−=3

sin3

cos8ππ

iz

⎥⎦

⎤⎢⎣

⎡−= iz

2

3

2

122

€

z = 2 − 6i

Multiplication of Complex Numbers

( ) ( )[ ]21212121 sincos θθθθ +++= irrzz

Find the Product if

⎟⎠

⎞⎜⎝

⎛ +=3

2sin

3

2cos21

ππiz & ⎟

⎠

⎞⎜⎝

⎛ +=6

11sin

6

11cos82

ππiz

=21zz ⎥⎦

⎤⎢⎣

⎡⎟⎠

⎞⎜⎝

⎛ ++⎟⎠

⎞⎜⎝

⎛ +6

11

3

2sin

6

11

3

2cos16

ππππi

⎥⎦

⎤⎢⎣

⎡ +=2

5sin

2

5cos16

ππi [ ] ii 16)1(016 =+=

2

π

Dividing Complex Numbers

( ) ( )[ ]21212

1

2

1 sincos θθθθ −+−= ir

r

z

z

Divide

( )oo iz 300sin300cos241 +=

( )oo iz 75sin75cos82 +=

[ ])75300sin()75300cos(8

24

2

1 oooo iz

z−+−=

[ ]oo i 225sin225cos3 +=

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−+⎟⎟

⎠

⎞⎜⎜⎝

⎛−=

2

2

2

23 i i

2

23

2

23−−=

2

1

z

z

DeMoivre’s Theorem and nth Roots

€

zn = r cosθ + isinθ( )[ ]n

( )θθ ninr n sincos +=

Ex. Find 12)31( i+−

Imagine how much fun it would be to multiply this example out 12 times. It would take forever. Using DeMovre’s Theorem, however, makes it short and simple.

First, convert to trigonometric form.

=+− 12)31( i12

3

2sin

3

2cos2 ⎥

⎦

⎤⎢⎣

⎡⎟⎠

⎞⎜⎝

⎛ +ππ

i

( ) ( ) ⎥⎦

⎤⎢⎣

⎡ +=3

212sin

3

212cos212 ππ

i ( )ππ 8sin8cos4096 i+=

( )i014096 += 4096=

nth Roots of a Complex Number

For a positive integer n, the complex number

z = r(cos x + i sin x)

has exactly n distinct nth roots given by

⎟⎠

⎞⎜⎝

⎛ ++

+n

ki

n

krn

πθπθ 2sin

2cos

where k = 0, 1, 2, . . . , n - 1

WATCH THE EXAMPLES CAREFULLY!!!

Ex. Find all the sixth roots of 1.

First, write 1 + 0i in trig. form.

1 + 0i is over 1 and up 0. Therefore, 1 is the hypotenuse and theta is 0o.

( ) 61

01 i+

€

1 cos0o + isin0o( )[ ]1

6

⎟⎠

⎞⎜⎝

⎛ ++

+=

6

20sin

6

20cos16 ππ k

ik

Now, we plug in 0, 1, 2, 3, 4, and 5 for k to find our six roots.

( ) 10sin0cos1 =+ i

ii2

3

2

1

3sin

3cos1 +=⎟

⎠

⎞⎜⎝

⎛ +ππ

ii2

3

2

1

3

2sin

3

2cos1 +−=⎟

⎠

⎞⎜⎝

⎛ +ππ

( ) 1sincos1 −=+ ππ i

ii2

3

2

1

3

4sin

3

4cos1 −−=⎟

⎠

⎞⎜⎝

⎛ +ππ

ii2

3

2

1

3

5sin

3

5cos1 −=⎟

⎠

⎞⎜⎝

⎛ +ππ

Find the three cube roots of z = -2+2i

Again, first convert to trig form.

€

−2 + 2i( )1

3 = 8 cos135o + isin135o( )[ ]1

3

€

81

2 ⎛ ⎝ ⎜ ⎞

⎠ ⎟1

3cos

135 + 360k

3+ isin

135 + 360k

3

⎛

⎝ ⎜

⎞

⎠ ⎟

For k = 0, 1, and 2, the roots are:

( )oo i 45sin45cos2 +

( )oo i 165sin165cos2 +

( )oo i 285sin285cos2 +