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Lesson 78 – Polar Form of Complex
Numbers
HL2 Math - Santowski
11/16/15
Relationships Among x, y, r, and 

x  r cos
y  r sin 
r  x2  y2
y
tan   , if x  0
x
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Polar Form of a Complex Number
 The expression
r (cos  i sin  )
is called the polar form (or trigonometric form) of the
complex number x + yi. The expression cos  + i sin  is
sometimes abbreviated cis .
Using this notation r (cos  i sin  ) is written r cis  .
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Example
 Express 2(cos 120 + i sin 120) in rectangular form.
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Example
 Express 2(cos 120 + i sin 120) in rectangular form.
1
cos120  
2
3
sin120 
2
 1
3
2(cos120  i sin120 )  2   , i

 2 2 
 1  i 3
 Notice that the real part is negative and the imaginary part is
positive, this is consistent with 120 degrees being a quadrant
II angle.
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Converting from Rectangular to Polar Form
 Step 1
Sketch a graph of the number x + yi in
the complex plane.
 Step 2
Find r by using the equation r  x 2  y 2 .
 Step 3
Find  by using the equation tan   xy , x  0
choosing the quadrant indicated in Step 1.
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Example
 Example: Find trigonometric notation for 1  i.
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Example
 Example: Find trigonometric notation for 1  i.
 First, find r.
r  a b
2
1  2
sin  

2
2
5

4
2
r  (1) 2  (1) 2
r 2
 Thus, 1  i  2  cos

1  2
cos 

2
2
5
5 
5
 i sin
or
2
cis

4
4 
4
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Product of Complex Numbers
 Find the product of
4(cos50  i sin 50 ) and 2(cos10  i sin10 ).
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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 .
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Example: Product
 Find the product of
4(cos50  i sin 50 ) and 2(cos10  i sin10 ).
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Example: Product
 Find the product of
4(cos50  i sin 50 ) and 2(cos10  i sin10 ).
 4(cos50  isin50 )    2(cos10  isin10 ) 

 

 4  2 cos(50  10 )  isin(50  10 ) 
 8(cos60  isin 60 )
1
3
 8  i 
2 
2
 4  4i 3
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Quotient of Complex Numbers
 Find the quotient.
16(cos 70  i sin 70 ) and 4(cos 40  i sin 40 )
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Quotient Theorem
 If
r1   cos1  i sin 1  and r2  cos 2  i sin  2 
are any two complex numbers, where
r2  cos 2  i sin  2  , r2  0,
then
r1  cos1  i sin 1 
r2  cos 2  i sin  2 
r1
 cos 1   2   i sin 1   2   .
r2
In compact form, this is written
r1 cis 1 r1
 cis 1   2 
r2 cis  2 r2
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Example: Quotient
 Find the quotient.
16(cos 70  i sin 70 ) and 4(cos 40  i sin 40 )
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Example: Quotient
 Find the quotient.
16(cos 70  i sin 70 ) and 4(cos 40  i sin 40 )

16(cos70  isin 70 ) 16
=
cos(70  40 )  isin(70  40 )
4
4(cos 40  isin 40 )

 4cos30  isin 30
 3 1 
 4
 i
 2 2 
 2 3  2i
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



If z  4 cos 40 o  isin40 o and w  6 cos120 o  isin120 o ,
find: (a) zw
(b) z w




If z  4 cos40 isin40 and w  6 cos120 isin120 ,
o
o
find: (a) zw
o
(b) z w

 
o

zw   4 cos 40  i sin 40   6 cos120  i sin120 

 


 4  6 cos 40 120  i sin 40 120
multiply the moduli

add the arguments (the i sine term
will have same argument)
 24  cos160   i sin160  
 24  0 .93969  0 .34202 i 
  22 . 55  8 . 21i
If you want the answer
in rectangular
coordinates simply
compute the trig
functions and multiply
the 24 through.




4 cos 40  isin40
z

w 6 cos120 o  isin120 o
 
o

o


4
o
o
o
o
 cos 40 120  isin 40 120
6
divide the moduli



subtract the arguments
    
2
 cos 80o isin 80o
3
    
2
o
o
 cos 280 isin 280
3
In rectangular
coordinates:
In polar form we want
an angle between 0
and 360° so add
360° to the -80°
2
 0 .1736  0 .9848i   0 .12  0 .66i
3
divide
Example
z1
z2
Where z1  3 2  3 2i  6  cos 45  i sin 45
z2  2 3  2i  4  cos 30  i sin 30 
Rectangular form
Trig form
divide
Example
z1
z2
Where z1  3 2  3 2i  6  cos 45  i sin 45
z2  2 3  2i  4  cos 30  i sin 30 
Rectangular form
Trig form
3 2  3 2i
2 3  2i
6  cos 45  i sin 45
4  cos30  i sin 30 
 3 2  3 2i   2 3  2i 



2
3

2
i
2
3

2
i



6 6  6 2i  6 6i  6 2i 2
12  4i 2
6 6  6 2i  6 6i  6 2
12  4
6
6 6 2
16
  6
6
cos  45  30   i sin  45  30  

4
3
 cos15  i sin15
2

6 6 2 i
16




 6 6 6 2 


1 
16
  15
  tan
 6 6 6 2 


216  72 12  72  216  72 12  72
16


r


2


 6 6 6 2   6 6 6 2 
 

r 
16
16

 


 

256
r
576
9 3


256
4 2
2
DeMoivre’s Theorem
Taking Complex Numbers to Higher Powers
z  r  cos  i sin  
z 2  ?? r  cos  i sin   r  cos  i sin  
 r  cos   2i cos sin   i sin  
2
2
2
 r 2  cos 2   sin 2   i 2sin  cos 
 r 2  cos 2  i sin 2 
z 2  r 2  cos 2  i sin 2 
z  r  cos  i sin  
z 2  r  cos  i sin   r  cos  i sin  
 r  cos   2i cos sin   i sin  
2
2
2
 r 2  cos 2   sin 2   i 2sin  cos 
 r 2  cos 2  i sin 2 
z 2  r 2  cos 2  i sin 2 
Nice 
What about
3
z ?
z  r  cos  i sin   , z 2  r 2  cos 2  i sin 2 
z 3  z  z 2  r 2  cos 2  i sin 2  r  cos  i sin  
 r 2  cos 2   2i cos sin   i 2 sin  
 r  cos   sin   i 2sin  cos 
2
2
2
 r  cos 2  i sin 2 
2
z 2  r 2  cos 2  i sin 2 
r 3  cos 2  i sin 2  cos  i sin   
r 3  cos 2 cos  i cos 2 sin   i sin 2 cos  i 2 sin 2 sin   
r 3  cos 2 cos  sin 2 sin   i cos 2 sin   i sin 2 cos  
r 3 cos 2 cos  sin 2 sin   i  sin 2 cos  cos 2 sin    
r 3 cos  2     i  sin  2       r 3  cos3  i sin 3 
z 3  r 3  cos3  i sin 3 
 Hooray!! 
We saw…
z 2  r 2  cos 2  i sin 2  and
z 3  r 3  cos3  i sin 3 
Similarly….
z 4  r 4  cos 4  i sin 4 
z 5  r 5  cos5  i sin 5 
z n  r n  cos n  i sin n 
We saw…
z 2  r 2  cos 2  i sin 2  and
z 3  r 3  cos3  i sin 3 
Similarly….
4
4
z  r  cos 4  i sin 4 
z 5  r 5  cos5  i sin 5 
z n  r n  cos n  i sin n 
Hooray DeMoivre and his incredible theorem
Powers of Complex Numbers
This is horrible in rectangular
form.
 a  bi 
 a  bi  a  bi  a  bi  ...  a  bi 
n
The best way to expand one
of these is using Pascal’s
triangle and binomial
expansion.
You’d need to use an i-chart
to simplify.
It’s much nicer in trig form. You just
raise the r to the power and multiply
theta by the exponent.
z  r  cos   i sin  
z n  r n  cos n  i sin n 
Example
z  5  cos 20  i sin 20 
z 3  53  cos 3  20  i sin 3  20 
z 3  125  cos 60  i sin 60 
De Moivre’s Theorem
 If
r1   cos1  i sin 1  is a complex number, and if n is
any real number, then
 r  cos1  i sin 1   r n  cos n  i sin n .
n
 In compact form, this is written
n
r
cis


r
 cis n .


n
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Example: Find (1  i)5 and express the result in rectangular
form.
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Example: Find (1  i)5 and express the result in rectangular
form.
 First, find trigonometric notation for 1  i
1  i  2  cos 225  i sin 225
 Theorem
 1  i 

5
  2  cos 225  i sin 225  
5
 2  cos(5  225 )  i sin(5  225 )
5
 4 2  cos1125  i sin1125

 2
2
 4 2
i

2 
 2
 4  4i
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
Example
 Let
z  1.  i
Find
z
10

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Example
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Further Examples
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nth Roots
 For a positive integer n, the complex number
a + bi is an nth root of the complex number x + yi if
 a  bi 
n
 x  yi.
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nth Root Theorem
 If n is any positive integer, r is a positive real number, and  is in
degrees, then the nonzero complex number r(cos  + i sin )
has exactly n distinct nth roots, given by
 where

n
r  cos  i sin   or
  360  k
n
n
r cis  ,

360  k
or  = 
, k  0,1,2,..., n  1.
n
n
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Example: Square Roots
 Find the square roots of 1
 3i
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Example: Square Roots
 Find the square roots of 1
 Polar notation: 1
 3i
 3i  2cos60  isin60
  60
 60
360 
360  
 2 cos60  isin60   2 cos   k 
 isin   k 

 


2
2
2
2





 2 cos 30  k 180  isin 30  k 180 


1
2
1
2

 For k = 0, root is
 For k = 1, root is





2 cos 210  isin 210
2 cos30  isin30
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Example: Fourth Root
 Find all fourth roots of 8  8i
3.
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Example: Fourth Root
 Find all fourth roots of 8  8i
rectangular form.
 Write in polar form.
3. Write the roots in
8  8i 3  16 cis 120
 Here r = 16 and  = 120. The fourth roots of this number
have absolute value 4 16  2.
120 360  k


 30  90  k
4
4
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Example: Fourth Root continued
 There are four fourth roots, let k = 0, 1, 2 and 3.
k 0
  30  90  0  30
k 1
  30  90  1  120
k 2
  30  90  2  210
k 3
  30  90  3  300
 Using these angles, the fourth roots are
2 cis 30 ,
2 cis 120 ,
2 cis 210 ,
2 cis 300
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Example: Fourth Root continued
 Written in rectangular form
3i
1  i 3
 3 i
1 i 3
 The graphs of the roots are all
on a circle that has center at the
origin and radius 2.
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Find the complex fifth roots of
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Find the complex fifth roots of
The five complex roots are:
for k = 0, 1, 2, 3,4 .
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Example
 Find all the complex fifth roots of 32
11/16/15
32  0i
2
2

32

0
r  a b
2
  tan 1
2
0
32
  tan 1 0
 322
 32
 0
32 cos  0   i sin  0  
5
  0 360k 
 0 360k  
32 cos  
  i sin  

5 
5 
5
 5
2 cos  72k   i sin  72k  
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2 cos  72k   i sin  72k  
k 0
2 cos  0   i sin  0    2 1  0i 
k 1
2 cos  72   i sin  72    .62  1.90i
k 2
2 cos 144   i sin 144    1.62 1.18i
k 3
2 cos  216   i sin  216    1.62 1.18i
k 4
2 cos  288  i sin  288    .62 1.90i
2
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Example
Find
3
i
You may assume it is the principle root you are
seeking unless specifically stated otherwise.
First express i as a complex number in standard form.
Then change to polar form
b
tan  
a
1
tan  
0
r  a 2  b2
1
  tan
0
1
0i
r 1
  90
1 cos  90   i sin  90  
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1 cos  90   i sin  90  
Since we are looking for the cube root, use DeMoivre’s Theorem
1 and raise it to the
power
3
 1

1

1 cos  90   i sin  90  

3

 3
1
3
1 cos  30   i sin  30  
 3
1
1 
i   3  1 i
2
2 2
 2
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Example:
Find the 4th root of 2  i
 2  i 
1
4
Change to polar form
5 cos 153.4   i sin 153.4  
Apply DeMoivre’s Theorem
 1

1
5 cos   153.4   i sin   153.4 
4
 4

1
4
1.22 cos  38.4   i sin  38.4  
.96  .76i
11/16/15