1.4. Polar Form for Complex Numbers 19

15. How would you define a complex version of the function log(1 + x)?

16. Verify that the power series defining arctan in Example 1.3.9 converges on the

disc D1(0).

1.4. Polar Form for Complex Numbers

If u is a complex number of modulus 1, then u = cos θ + i sin θ = eiθ, where θ is

an angle from the positive x axis to the ray from the origin through u. This angle

is measured in radians and is the signed length of an arc on the unit circle joining

(1, 0) to u. It is positive if the arc is traversed in the counterclockwise direction

and negative if it is traversed in the clockwise direction.

If z is any non-zero complex number, then z/|z| is a complex number of modulus

1 and so it has the form eiθ for some θ. If we set r = |z|, then

z = r

eiθ

.

This is the polar form of the complex number z. The angle θ is called the argument

of z and is denoted arg z.

Comments:

(1) By Euler’s identity (Theorem 1.3.4), r

eiθ

= r(cos θ + i sin θ).

(2) The number z must be non-zero for the argument to be defined. The polar

form of z = 0 is just z = 0.

(3) For z = 0, there are infinitely many angles θ for which z = r

eiθ.

Given one

such θ the others are all of the form θ + 2πn, where n is an integer. Thus,

arg(z) is not a single number, but a collection of numbers that differ from

one another by multiples of 2π. We can specify a particular one of these by

insisting it lie in a particular half-open interval of length 2π – such as [0, 2π)

or (−π, π].

(4) The numbers (r, θ), where r = |z| and θ is a value of arg(z), are polar coordi-

nates for z = x + iy considered as a point (x, y) in

R2.

(5) We now have two useful ways of expressing a complex number z, the standard

form x + iy and the polar form r

eiθ.

It is important to be able to easily

change from one to the other.

Example 1.4.1. Find the polar form for the complex number z = 2 + 2i.

Solution: We have

r = 22 + 22 = 2

√

2 and θ = arctan(1) = π/4.

Thus, z = 2

√

2 eπi/4 is the polar form of z.

Example 1.4.2. Put the complex number z = 2

eπi/6

in standard form x + iy.

Solution: We have

x = 2 cos π/6 =

√

3 and y = 2 sin π/6 = 1

and so z =

√

3 + i.