jeff bivin -- lzhs trig (polar) form of a complex number by: jeffrey bivin lake zurich high school...
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Jeff Bivin -- LZHS
Trig (Polar) Form of a Complex Number
By: Jeffrey BivinLake Zurich High School
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