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A FIRST COURSE
IN LINEAR ALGEBRA
An Open Text by Ken Kuttler
Complex Numbers - Polar Form
Lecture Notes by Karen Seyffarth∗
Adapted by
LYRYX SERVICE COURSE SOLUTION
∗
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Complex Numbers - Polar Form
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Complex Numbers in Polar Form
√
Suppose z = a + bi, and let r = |z| = a2 + b 2 . Then r is the distance
from z to the origin. Denote by θ the angle that the line through 0 and z
makes with the positive x-axis (measured clockwise).
y
z = a + bi
r
b
θ
0
a
x
Then θ is an angle defined by cos θ = ar and sin θ = br , so
z = r cos θ + r sin θi = r (cos θ + i sin θ).
θ is called an argument of z, and is denoted arg z.
Complex Numbers - Polar Form
Polar Form
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Definition (Polar Form of a Complex Number)
Let z be a complex number with |z| = r and arg z = θ. Then
z = re iθ = r (cos θ + i sin θ)
is called a a polar form of z. Since arg z is not unique, we do not write the polar
form of z.
Definition
Let z be a complex number with |z| = r . The principal argument of z is the
unique angle θ = arg z (measured in radians) such that
−π < θ ≤ π.
Example
Let z = 1. To convert
z to polar form, we need to find r and θ so that 1 = re iθ .
√
2
Now r = |z| = 1 = 1, and θ = 0 is an argument for z = 1. However, we may
also write
1 = e 2πi , 1 = e −2πi , e 4πi , e 6πi , . . .
Since sine and cosine have periodicity 2π, we may add (or subtract) multiples of
2π to any argument.
Complex Numbers - Polar Form
Polar Form
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Example (Converting to Polar Form)
√
Let z =√−2 + 2 3i. To convert z to polar form, we need to find r and θ so that
−2 + 2 3i = re iθ . Since r = |z|,
q
p
√
√
r = (−2)2 + (2 3)2 = 4 + 4(3) = 16 = 4.
√
There are two approaches to finding an argument, θ. One is to graph −2 + 2 3
in the complex plane.
y
√
(−2, 2 3)
√
2 3
θ
4
2
0
x
The
√ triangle sitting on the negative half of the real axis has sides of length 2,
2 3, and 4; you should recognize this as a right triangle whose other two angles
measure π3 and π6 . From this, we see that θ = 2π
3 is an argument of z.
Therefore, z can be written in polar form as
z = 4e i(2π/3) .
Complex Numbers - Polar Form
Polar Form
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Example (Converting to Polar Form – continued)
√
The other approach to finding an argument, θ, for z = −2 + 2 3i is as follows.
We’ve already calculated |z| = r = 4. By definition, θ is an angle satisfying
√
√
1
2 3
3
−2
= − and sin θ =
=
.
cos θ =
2
4
2
4
By graphing the point (− 12 ,
be written in polar form as
√
3
2 ),
we again determine that θ =
z = 4e i(2π/3) .
Complex Numbers - Polar Form
Polar Form
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2π
3 ,
and thus z can
Exercises
Convert each of the following complex numbers to polar form.
1
2
3
4
3i = 3e (π/2)i .
√
√
−1 − i = 2e −(3π/4)i = 2e (5π/4)i .
√
3 − i = 2e −(π/6)i .
√
√
3 + 3i =2 3e (π/3)i .
Complex Numbers - Polar Form
Polar Form
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Converting from Polar Form to Cartesian form
Problem
Let z = 2e 2πi/3 . Write z in the form z = a + bi (this is called Cartesian
form or Standard form).
Solution
First, remember that e iθ = cos θ + i sin θ, and thus
e 2πi/3 = cos(2π/3) + i sin(2π/3)
√
1
3
= − +i
.
2
2
Therefore
z = 2e
Complex Numbers - Polar Form
2πi/3
√ !
1
3
= 2 − +i
2
3
√
= −1 + 3i.
Polar Form
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Exercises
Express each of the following complex numbers in Cartesian form.
1
2
3
3e −πi = −3
√
√
2e 3πi/4 = − 2 + i 2
√
√
2 3e −2πi/6 = 3 − 3i
Complex Numbers - Polar Form
Polar Form
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Multiplication - again!
Problems involving multiplication of complex numbers can often be solved more
easily by using polar forms of the complex numbers.
Theorem
If z1 = r1 e iθ1 and z2 = r2 e iθ2 are complex numbers, then
z1 z2 = r1 r2 e i(θ1 +θ2 )
Theorem (De Moivre’s Theorem)
If θ is any angle and n is a positive integer,
n
e iθ = e inθ .
As an immediate consequence of De Moivre’s Theorem, we have that for any real
number r > 0 and any positive integer n,
(re iθ )n = r n e inθ .
Complex Numbers - Polar Form
Polar Form
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Problem
√
Express (1 − i)6 ( 3 + i)3 in the form a + bi.
Solution
Let z = 1 − i =
compute z 6 w 3 .
√
2e −πi/4 and w =
√
3 + i = 2e πi/6 . We want to
√
z 6 w 3 = ( 2e −πi/4 )6 (2e πi/6 )3
= (23 e −6πi/4 )(23 e 3πi/6 )
= (8e −3πi/2 )(8e πi/2 )
= 64e −πi
= 64e πi
= 64(cos π + i sin π)
= −64.
Complex Numbers - Polar Form
Polar Form
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Problem
Express
1
2
√
3
2 i
−
17
in the form a + bi.
Solution
Let z =
1
2
√
−
3
2 i
= e −πi/3 . Then
z 17 =
e −πi/3
17
= e −17πi/3
= e πi/3
π
π
= cos + i sin
3√
3
1
3
=
+
i.
2
2
Complex Numbers - Polar Form
Polar Form
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