Jeff Bivin -- LZHS Trig (Polar) Form of a Complex Number By: Jeffrey Bivin Lake Zurich High School [email protected] Last Updated: February 23, 2011.

Jeff Bivin -- LZHS

Trig (Polar) Form of a Complex Number

By: Jeffrey BivinLake Zurich High School

[email protected]

Last Updated: February 23, 2011

Jeff Bivin -- LZHS

Absolute Value of a Complex Number

a bi

The distance from the origin to the point (a, b).

2 23 2 3 2 13i

2 25 3 ( 5) 3 34i

(a, b)

a

b

2 2a bi a b

Jeff Bivin -- LZHS

Trig (Polar) form of a Complex Number

cosa

r

(a, b)

2 2:where r a b

cos sina bi r r i

a

rb sin

b

r

cosr a sinr b

a bi

cos sina bi r i

Jeff Bivin -- LZHS

Convert 1 - i to Trig(Polar) Form2 21 ( 1) 2r 1

1tan

4

4 41 1 2 cos sini i

Jeff Bivin -- LZHS

Convert -1 + i to Trig(Polar) Form2 2( 1) 1 2r 1

1tan 34

3 34 41 1 2 cos sini i

Select the correct

Quadrant

Jeff Bivin -- LZHS

Convert 3 - 4i to Trig(Polar) Form2 23 ( 4) 5r 4

3tan

53.130

3 4 5 cos 53.130 sin 53.130i i

Jeff Bivin -- LZHS

Convert 5(cosπ + i·sinπ ) to Standard Form

5 1 0i

5

1 1(cos sin )z r i

Jeff Bivin -- LZHS

Convert to Standard Form

3 12 23 i

3 3 32 2 i

5 56 63 cos sini

Jeff Bivin -- LZHS

Multiplication of Complex Numbersin Trig(Polar) Form

1 1(cos sin )z r i 2 2 (cos sin )z r i

1 2 1 2(cos sin ) (cos sin )z z r i r i

1 2 (cos sin )(cos sin )r r i i

21 2 cos cos cos sin sin cos sin sinr r i i i

1 2 cos sinr r i

1 2 cos cos sin sin sin cos cos sinr r i

-1

Jeff Bivin -- LZHS

Multiplication of Complex Numbersin Trig(Polar) Form

1 3 35 cos sinz i 5 52 6 63 cos sinz i

5 51 2 3 6 3 65 3 cos sinz z i

7 71 2 6 615 cos sinz z i

3 11 2 2 215z z i

15 3 151 2 2 2z z i

Jeff Bivin -- LZHS

Division of Complex Numbersin Trig(Polar) Form

1 1(cos sin )z r i 2 2 (cos sin )z r i

11 2

2

cos sinr

z z ir

1

1 2

2

cos sin

cos sin

r iz z

r i

1

2

cos sin cos sin

cos sin cos sin

r i i

r i i

21

2 2 22

cos cos cos sin sin cos sin sin

cos sin

r i i i

r i

1

2 22

cos cos sin sin sin cos cos sin

cos sin

r i

r

Jeff Bivin -- LZHS

Division of Complex Numbersin Trig(Polar) Form

1 3 35 cos sinz i 5 52 6 63 cos sinz i

5 5 51 2 3 3 6 3 6cos sinz z i

51 2 3 2 2cos sinz z i

51 2 3 0 ( 1)z z i

51 2 3z z i

Jeff Bivin -- LZHS

Powers of Complex Numbersin Trig(Polar) Form

(cos sin )z r i 2 (cos sin ) (cos sin )z r i r i 2 2 (cos 2 sin 2 )z r i

3 2 (cos 2 sin 2 ) (cos sin )z r i r i 3 3(cos3 sin 3 )z r i

(cos sin )n nz r n i n

This is called DeMoivre’s

Theorem

Jeff Bivin -- LZHS

Use DeMoivre’s Thm to evaluate: 103 3 i

1010 5 5

6 62 3 cos 10 sin 10z i

Convert to trig form: 3 3z i 223 3 12 2 3r

33tan 56

10 25 253 3248832 cos sinz i

103 3248832 cos sinz i

10 312 2248832z i

10 124416 124416 3z i

Jeff Bivin -- LZHS

Solve: x6 – 1 = 0

3 31 1 0x x

2 21 1 1 1 0x x x x x x

2 21 0 1 0 1 0 1 0x x x x x x

1x

21 1 41121x

1 32ix 1x

21 ( 1) 4 11

21x

1 32ix

This one worked nicely because the equation was factorable. What

happens if it isn’t nicely factorable?

Just for Fun

Jeff Bivin -- LZHS

Roots of Complex Numbersin Trig(Polar) Form

(cos sin )z r i

1 12 2cos sinn nn k kn nz z r i

DeMoivre’s Theorem will help

us with this

0, 1, 2, , 1where k n

Jeff Bivin -- LZHS

Find the 6 roots of 1i.e. x6 = 1

1 1 0i

16 0 2 0 2

6 61 cos sink ki

0, 1, 2, 3, 4, 5where k

6 1x

2 21 0 1r 01tan 0

12 2cos sinn k kn nr i

Jeff Bivin -- LZHS

16 0 2 0 2

6 61 cos sink ki 0, 1, 2, 3, 4, 5where k

160 : 1 cos0 sin 0 1 0 1k i i 16 32 2 1

6 6 3 3 2 21: 1 cos sin cos sink i i i 16 34 4 2 2 1

6 6 3 3 2 22 : 1 cos sin cos sink i i i 16 6 6

6 63: 1 cos sin cos sin 1 0 1k i i i 16 38 8 4 4 1

6 6 3 3 2 24 : 1 cos sin cos sink i i i 16 310 10 5 5 1

6 6 3 3 2 25 : 1 cos sin cos sink i i i

Jeff Bivin -- LZHS

Find the 3 cube roots of -8

8 0z i

13 2 2

3 38 cos sink ki

0, 1, 2where k

3 8x 2 28 0 8r

08tan

12 2cos sinn k kn nr i

Jeff Bivin -- LZHS

13 2 2

3 38 cos sink ki 0, 1, 2where k

13 31

3 3 2 20 : 8 cos sin 2 1 3k i i i

13 3 3

3 31: 8 cos sin 2 1 0 2k i i

13 35 5 1

3 3 2 22 : 8 cos sin 2 1 3k i i i

Jeff Bivin -- LZHS

Find the 4 roots of 3 - 4i

3 4z i

14 53.130 360 53.130 360

4 45 cos sink ki

0, 1, 2, 3where k

4 3 4x i

2 23 ( 4) 5r 43tan 53.130

12 2cos sinn k kn nr i

Jeff Bivin -- LZHS

0, 1, 2, 3where k

141: 5 cos76.7175 sin 76.7175 .344 1.455o ok i i 142 : 5 cos166.7175 sin166.7175 1.455 0.344o ok i i 143: 5 cos 256.7175 sin 256.7175 .344 1.455o ok i i

140 : 5 cos 13.283 sin 13.283 1.455 0.344o ok i i

14 53.130 360 53.130 360

4 45 cos sink ki

Jeff Bivin -- LZHS

Find the 15 roots of 4 - 7i

Just Kidding!!!!!

0 : 1.146 0.081k i 1: 1.080 0.393k i 2 : 0.827 0.798k i 3: 0.431 1.065k i 4 : 0.040 1.149k i 5 : 0.504 1.033k i 6 : 0.880 0.739k i 7 : 1.105 0.317k i

8 : 1.138 0.160k i 9 : 0.975 0.609k i

10 : 0.643 0.953k i 11: 0.200 1.132k i 12 : 0.278 1.115k i 13: 0.707 0.906k i 14 : 1.015 0.540k i