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7.5 Trigonometric (Polar) Form of
Complex Numbers
• The Complex Plane and Vector Representations
Call the horizontal axis the real
axis and the vertical axis the
imaginary axis. Now complex
numbers can be graphed in this
complex plane.
The sum of two complex numbers
can be represented graphically by
the vector that is the resultant of
the sum of vectors corresponding
to the two numbers.
Copyright © 2007 Pearson Education, Inc.
Slide 10-1
7.5 Expressing the Sum of Complex
Numbers Graphically
Example Find the sum of 6 – 2i and –4 – 3i. Graph
both complex numbers and their resultant.
Solution
(6 – 2i) + (–4 – 3i) = 2 – 5i
Copyright © 2007 Pearson Education, Inc.
Slide 10-2
7.5 Trigonometric (Polar) Form
The graph shows the complex
number x + yi that corresponds
to the vector OP.
Relationship Among x, y, r, and 
y  r sin 
x  r cos 
y
2
2
tan   , if x  0
r x y
x
Copyright © 2007 Pearson Education, Inc.
Slide 10-3
7.5
Trigonometric (Polar) Form
• Substituting x = r cos  and y = r sin  into x + yi
gives
x  yi  (r cos  )  (r sin  )i
 r (cos   i sin  ).
Trigonometric or Polar Form of a Complex Number
The expression
r(cos  + i sin  )
is called the trigonometric form or polar form of the
complex number x + yi.
Copyright © 2007 Pearson Education, Inc.
Slide 10-4
7.5
Trigonometric (Polar) Form
• Notation:
cos  + i sin  is sometimes written cis  . Using
this notation, r(cos  + i sin  ) is written r cis .
• The number r is called the modulus or absolute
value of the complex number x + yi.
• Angle  is called the argument of the complex
number x + yi.
Copyright © 2007 Pearson Education, Inc.
Slide 10-5
7.5 Converting from Trigonometric Form
to Rectangular Form
Example Express 2(cos 300º + i sin 300º) in
rectangular form.
Analytic Solution
3
1
2(cos 300  i sin 300 )  2  i   1  i 3.
2 
2


Graphing Calculator Solution
Copyright © 2007 Pearson Education, Inc.
Slide 10-6
7.5 Converting from Rectangular to
Trigonometric Form
Converting from Rectangular to Trigonometric
Form
1. Sketch a graph of the number in the complex
plane.
2. Find r by using the equation
x2  y 2 .
3. Find  by using the equation tan  = y/x, x  0,
choosing the quadrant indicated in Step 1.
Copyright © 2007 Pearson Education, Inc.
Slide 10-7
7.5 Converting from Rectangular to
Trigonometric Form
Example Write each complex number in trigonometric
form.
(a )  3  i
(b)  3i
Solution
(a) Start by sketching the graph
of  3  i in the complex
plane.
Then find r.
r  x2  y 2
 ( 3) 2  12  2
Copyright © 2007 Pearson Education, Inc.
Slide 10-8
7.5 Converting from Rectangular to
Trigonometric Form
Now find .
 is in quadrant II and tan  =  33 ,
the reference angle in quadrant II is 6 .
y
tan  
x
3
 5


    
 3
3
6 6
1
Therefore, in polar form,
5
5 
5

 3  i  2 cos
 i sin   2 cis .
6
6 
6

Copyright © 2007 Pearson Education, Inc.
Slide 10-9
7.5 Converting from Rectangular to
Trigonometric Form
 3i  0  3i
(b)
r  0   3  3
2
2
y
tan   is undefined, so use a
x
different way to determine .
From the graph,  = 270º. In trigonometric form,
 3i  3(cos 270  i sin 270 )  3 cis 270.
Copyright © 2007 Pearson Education, Inc.
Slide 10-10
7.5 Deciding Whether a Number is in the
Julia Set
Example The fractal called the Julia set is shown in the figure.
To determine if a complex number z = a + bi is in this Julia set,
perform the following sequence of calculations. Repeatedly
compute the values of z2 – 1, (z2 – 1)2 –1, [(z2 – 1)2 –1]2 – 1, . . . .
If the moduli of any of the resulting complex numbers exceeds 2,
then z is not in the Julia set. Otherwise z is part of this set and the
point (a, b) should be shaded in the graph.
Copyright © 2007 Pearson Education, Inc.
Slide 10-11
7.5 Deciding Whether a Number is in the
Julia Set
Determine if z = 0 + 0i belongs to the Julia set.
z  0  0i  0
Solution
So,
z 2  1  02  1  1
( z 2  1) 2  1  (1) 2  1  0
[( z  1)  1]  1  0  1  1
2
2
2
2
and so on. The calculations repeat as 0, –1, 0, –1, and so
on. The moduli are either 0 or 1, therefore, 0 + 0i belongs to the
Julia set.
Copyright © 2007 Pearson Education, Inc.
Slide 10-12
7.5 Products of Complex Numbers in
Trigonometric Form
• Multiplying complex numbers in rectangular form.
1i 3  2
3  2i   2 3  2i  6i  2 3  4 3  4i
• Multiplying complex numbers in trigonometric form.
1i 3  2
3  2i   2(cos 60  i sin 60 )   4(cos150 i sin 150 ) 
 2  4(cos 60 cos150  i sin 60 cos150




 i cos 60 sin 150  i sin 60 sin 150 )



2

 8[(cos 60 cos150  sin 60 sin 150 )




 i (sin 60 cos150  cos 60 sin 150 )]




 8[cos( 60  150 )  i sin( 60  150 )]




 8(cos 210  i sin 210 )

Copyright © 2007 Pearson Education, Inc.

Slide 10-13
7.5 Products of Complex Numbers in
Trigonometric Form
Product Theorem
If r1 (cos 1  i sin 1 ) and r2 (cos  2  i sin  2 ) are any two
complex numbers, then
r1 (cos 1  i sin 1 )  r2 (cos  2  i sin  2 )
 r1r2 cos(1   2 )  i sin(1   2 ).
In compact form, this is written
(r1 cis 1 )(r2 cis  2 )  r1r2 cis (1   2 ).
Copyright © 2007 Pearson Education, Inc.
Slide 10-14
7.5 Using the Product Theorem
Example Find the product of 3(cos 45º + i sin 45º) and
2(cos 135º + i sin 135º).
Solution
[3(cos 45  i sin 45 )][ 2(cos 135  i sin 135 )]
 3  2[cos( 45  135 )  i sin( 45  135 )]
 6(cos 180  i sin 180 )
 6(1  i  0)
 6
Copyright © 2007 Pearson Education, Inc.
Slide 10-15
7.5 Quotients of Complex Numbers in
Trigonometric Form
• The rectangular form of the quotient of two complex numbers.
1 i 3
(1 i 3)(2 3  2i )
 2 3  2i  6i  2i 2 3


 2 3  2i ( 2 3  2i )(2 3  2i )
12  4i 2
8i
1

 i
16
2
• The polar form of the quotient of two complex numbers.


1  i 3  2 cos i sin 

3
3
5
5 
 2 3  2i  4 cos  i sin


6
6 
1
1   
 

 i  cos    i sin    
2
2  2
 2 
Copyright © 2007 Pearson Education, Inc.
Slide 10-16
7.5 Quotients of Complex Numbers in
Trigonometric Form
Quotient Theorem
If r1(cos 1 + i sin 1) and r2(cos 2 + i sin 2) are
complex numbers, where r2(cos 2 + i sin 2)  0, then
r1 (cos 1  i sin 1 ) r1
 cos(1   2 )  i sin(1   2 ).
r2 (cos  2  i sin  2 ) r2
In compact form, this is written
r1 cis 1 r1
 cis (1   2 )
r2 cis  2 r2
Copyright © 2007 Pearson Education, Inc.
Slide 10-17
7.5 Using the Quotient Theorem
Example Find the quotient
rectangular form.
Solution
10 cis ( 60 )
5 cis (150 )
. Write the result in
10 cis (60 ) 10


 cis (60  150 )

5 cis 150
5
 2 cis (210 )
 2cos( 210 )  i sin( 210 )
 3  1 
 2 
 i 
 2 
 2
  3i
Copyright © 2007 Pearson Education, Inc.
Slide 10-18