MATH 117
The Polar Form of
Complex Numbers
At the end of the 18th century, Caspar Wessel (1745 – 1818), Jean Robert Argand (1768 –
1822), and Carl Friedrich Gauss (1777 – 1855) independently came up with a geometric
interpretation of a complex number z = a + bi .
The complex number could simply be considered as an ordered pair ( a , b ) in the
x y -plane. Then the point ( a , b ) could be further represented in polar form by an angle
θ and a radius r .
The angle θ is always measured counterclockwise from
the positive x -axis with 0 ≤ θ < 2π. The radius is given by
r
θ
r = a 2 + b2 , and we always take r ≥ 0.
The Polar Form
First recall Euler’s Formula which states that e i x = cos x + i sin x . Now because cos θ =
adj
opp
a
b
= , then a = r cos θ . Likewise since sin θ =
= , then b = r sin θ . Thus, z =
hyp
hyp
r
r
a + bi = r cos θ + (rsin θ )i = r (cos θ + i sin θ ) , and so by Euler’s Formula:
z = r e iθ
Conversion: Polar to Rectangular
Given the radius and angle ( r , θ ), we can easily convert back to rectangular form z =
a + bi since a = r cos θ and b = r sin θ .
Example 1. Convert to rectangular form
(a) z = 4 e 5 π i / 3
(b) z = 10e 7 π i / 6 .
Solution. (a) Here r = 4 and θ = 5π/3, so a = 4 cos(5π / 3) = 2 and b = 4 sin(5π / 3) = –
2 3 . Thus, z = 2 − 2 3 i .
− 3
 + 10 ×  −1  i = −5 3 − 5 i .
(b) z = 10cos(7π / 6) + 10sin(7π / 6) i = 10 × 
 2
 2 
To obtain the above results in decimal form with your calculator, simply enter
4 ^(5π /3). (Make sure that you are in radian mode.)
Conversion: Rectangular to Polar
Given z = a + bi , we also can find the radius and angle ( r , θ ).
2
2
r = a + b . Also tan θ = b / a . To find θ , compute tan
appropriately.
−1
We know that
(b / a) and adjust the quadrant
π
π
−1
< tan −1 x < . By definition, tan x is the angle θ between –π/2 and
2
2
−1
π/2 such that tan θ = x . For instance tan 1 = π/4 becasue tan(π / 4) = 1.
Recall that −
f (x) = tan x
Domain: –π/2 < x < π/2
Range: –∞ < y < ∞
x
π
2
π
–
3
π
–
4
π
–
6
–
0
π
6
π
4
π
3
π
2
tan x
−1
f (x) = tan x
Domain: –∞ < x < ∞
Range: –π/2 < y < π/2
x
–∞
–∞
– 3
– 3
–1
1
–
3
–1
1
–
3
0
1
3
0
1
3
1
1
3
3
∞
∞
tan −1 x
π
2
π
–
3
π
–
4
π
–
6
–
0
π
6
π
4
π
3
π
2
Note that angles in the 4th quadrant are written as negative angles. However, for
−1
the angle θ in the polar form of z , we want 0 ≤ θ < 2π; therefore we must adjust tan x
by adding π or 2π appropriately.
Example 2. Find the polar form of the complex numbers.
(a) z = −3 3 + 3i
(b) z = −6 − 6i
(c) z = 4 − 4 3 i
Solution. (a) The point ( −3 3 , 3 ) is in the 2nd quadrant. The
2
2
radius is r = (−3 3 ) + 3
=
36 = 6. Next b / a = −1 / 3 and
−1
tan (−1 / 3) = –π/6. So we add π to obtain the angle in the
2nd quadrant. Thus θ = 5π/6 and z = 6e 5 π i / 6 .
(b) The point (–6, –6) is in the 3rd quadrant. The radius is
r = (−6)2 + (−6)2 = 6 2 . Next b / a = 1 and tan −1 (1) = π/4.
But we add π to obtain the angle in the 3rd quadrant. Thus θ =
5π/4 and z = 6 2 e 5π i / 4 .
(c) The point (4, − 4 3 ) is in the 4th quadrant. The radius is
r = 4 2 + (−4 3) 2 = 64 = 8. Next b / a = – 3 and tan −1 (− 3)
= –π/3. But we now add 2π to obtain an angle θ in the 4th
quadrant with 0 ≤ θ < 2π. Thus θ = 5π/3 and z = 8e 5 π i / 3 .
Most often though, we do not have complex numbers that give one of the special
unit circle angles. In these cases, we usually write the angle in degree form.

For example, if z = −4 + 2i , then r = 20 and tan −1 (b / a) = tan −1 (−1 / 2) ≈ −26.565

(using Degree mode on calculator). So in the 2nd quadrant, θ ≈ 153.435 .
The Axes Angles
Consider the complex numbers 8, 6i , –10, – 20i . In the x y plane, these values are
points on the axes: (8, 0), (0, 6), (–10, 0), and (0, –20). The radii are the positive values 8,
6, 10, and 20, and the angles are 0, π/2, π, and 3π/2. Thus, by Euler’s Formula, we have
8 = 8e 0 i , 6i = 6 e π i / 2 , –10 = 10 e π i , and – 20i = 20e 3 π i / 2 .
Geometric Interpretations
Powers: Let z = r e i θ be the polar form of a complex number. Then we can compute
various integral powers of z and see the effect on the radius and the angle. Since
(
zn = r e i θ
)
n
= rn e i (n θ ) ,
n
we see that z raises the radius to the n th power and rotates the angle by a factor of n .



2
3
For example, if z = 2 e i (20 ) , then z = 4 e i (40 ) , z = 8e i (60 ) , and so on.
2
z , z , and z
3
Reciprocal: If z = r e i θ , then z−1 = r −1 e i (− θ ) . Thus, we obtain the reciprocal of the
radius and we have the same the angle but rotated in the negative direction.


1
For example, if z = 3e i (45 ) , then z−1 = e i (− 45 ) .
3
z and z
iθ
−1
iθ
Products and Quotients: Let z1 = r1 e 1 and z2 = r2 e 2 . Then
i (θ + θ )
iθ
iθ
z1 × z2 = r1 e 1 × r2 e 2 = ( r1 r2 ) e 1 2 .
For a product, we multiply the radii and

add the angles. If z1 = 2 e i (20 ) and z2 =


3e i (45 ) , then z1 × z2 = 6e i (65 ) .
z1 and z2
Also
iθ
z1
r i (θ − θ )
r e 1
= 1 iθ = 1 e 1 2 .
z2
r2
r2 e 2
Thus for a quotient, we divide the radii and subtract the angles.
z1 × z2
Exercises
1. Find the rectangular form of the complex numbers.
(a) z = 8e π i
(b) z = 4 e πi / 3
(c) z = 12e 11π i / 6
2. Convert the complex numbers to the polar form r e i θ .
(a) z = −5 2 + 5 2 i
(b) z = −3 − 3 3 i
(c) 7 − 16 i
3. Let z1 = 4 e πi / 3 and z2 = −3 − 3 3 i . (a) Multiply in rectangular form. (b) Multiply
in polar form. (c) Convert the polar product to rectangular form to verify that the
result is the same as in (a).