Proposition 1.2.16: Finding Roots

For any positive integer n and any non-zero complex number a = r cis(t) the equation zn = a has exactly n distinct roots given by:
z =
where k = 0, 1, 2, ... n-1.
Context Context

As is often the case in mathematics, the proof is easy if you happen to know the correct answer by working backwords. In our case we start by saying: let a = r cis(t) and define

z =
Then, according to our theorem on muliplying geometrically we have:

for all k = 0, 1, 2, ... n-1 because of the periodicity of cos.

[ x ]

It is more enlightning to try to understand how this theorem works geometrically. For simplicity take a = cis(t), i.e. |a|=1:

To find an n-th root of a vector with angle t, divide that angle by n

That gives you the first root.

Divide the unit circle into n equally spaced pieces, starting at the first angle t/n

That will be your n-roots total. As an example, let's find the three third-roots of i, i.e. we want to find all solutions to z3 = i.

Step 1: Draw the vector i

Step 2: Divide the angle by 3 for your first root

Step 3: Draw 3 equally spaced segments, starting at the first root

Interactive Complex Analysis, ver. 1.0.0
(c) 2006-2007, Bert G. Wachsmuth
Page last modified: May 29, 2007