Mastering Mathematics Smartly by Wee Wen Shih

A unique self-help website that provides comprehensive coverage of mathematics at A-level & beyond, written in a student-friendly style.

Roots of unity

A typical problem involves these steps:

(1) Make the right-hand-side 1.

(2) Make the left-hand-side into an expression that is raised to a power.

(3) Use a suitable substitution and the general solution of z.

(4) Solve for z.

Step (4) usually requires some effort in simplifying. Knowledge of rules of indices (if z is to be expressed in exponential form) and trigonometric identities (if z is to be expressed in polar form) will be necessary.

See example.

Try this!

To understand the steps of finding the n th root of a complex number, please refer to this link. Focus on "Theorem A.5" on page F9 and go over "Example 10" on page F10.

e i theta and roots

One common question involves an expression in exponential form and its application in finding roots of an equation.

Refer to this example to see what I mean.

Form of solutions

We recall that

z^n = re^{\text{i}\theta} \Rightarrow z = r^{\frac{1}{n}}e^{\text{i}\left(\frac{\theta + 2k\pi}{n}\right)}, k = 0, 1, 2, ..., n - 1.
 
Students often wonder whether k can be be written as k = 0, \pm1, \pm2, ...?
 
Let's consider three examples to understand this further.
 
Example 1: Solve z^3 = 1.
 
Firstly, notice that the cubic equation has only real coefficients.
 
We obtain z = e^{\frac{2k\pi}{3}\text{i}}, k = 0, 1, 2 via the above result.
 
If we were to substitute values of k, we have the roots z = 1, e^{\frac{2\pi}{3}\text{i}}, e^{\frac{4\pi}{3}\text{i}}.
 
Since arguments are to be within the principal range, we re-express the roots to give z = 1, e^{\pm\frac{2\pi}{3}\text{i}}. So k can take values 0, \pm1.
 
Example 2: Solve z^3 = \text{i}.
 
Note that there is a complex constant.
 
We obtain z = e^{\frac{\left(\frac{\pi}{2} + 2k\pi \right)}{3}\text{i}}, k = 0, 1, 2 by the same result.
 
The roots are e^{{\frac{\pi}{6}}\text{i}}, e^{{\frac{5\pi}{6}}\text{i}}, e^{-{\frac{\pi}{2}}\text{i}} corresponding to k taking values 0, 1, 2.

We can also obtain the same roots by substituting k with values 0, \pm1. In particular, when k =-1, we obtain the root e^{-\frac{\pi}{2}\text{i}}.

Example 3: Solve z^4 = \text{i}.

Note that we have an even power.

We obtain z = e^{\frac{(\frac{\pi}{2}+2k\pi)}{4}\text{i}}, k = 0, 1, 2, 3.

We can also write k to take values 0, \pm1, 2. One should verify that both sets of k values give us the same roots. In particular, k = -1 and k = 3 give the same root.