Chapter 6.6, Problem 64E

Solution

To prove: That the $n\text{th}$ roots of the complex number are spaced $\frac{2\pi }{n}$ radians apart on a circle with radius $\sqrt[n]{r}$ .

Given information:

The complex number is $z=r\left(\cos \theta +i\sin \theta \right)$

Formula used:

The $n\text{th}$ roots of a complex number $z=r\left(\cos \theta +i\sin \theta \right)$ which are $n$ distinct complex numbers are given by,

  $\sqrt[n]{r}\left(\cos \frac{\theta +2\pi k}{n}+i\sin \frac{\theta +2\pi k}{n}\right)$

Here, $k=0,1,2,\mathrm{...},n-1$

Calculation:

In this polar form, the modulus or absolute value is $\sqrt[n]{r}$ which the distance of the root from the origin is. Since all roots have the same modulus, then they all lie on a circle with radius $\sqrt[n]{r}$ .

Using any two consecutive values of $k$ , say $k$ and $k+1$ , the difference between two roots is:

  $\begin{array}{c}\frac{\theta +2\pi \left(k+1\right)}{n}-\frac{\theta +2\pi k}{n}=\frac{\theta +2\pi \left(k+1\right)-\theta -2\pi k}{n}\\ =\frac{\theta +2\pi k+2\pi -\theta -2\pi k}{n}\\ =\frac{2\pi }{n}\end{array}$

Therefore, it is proved that the $n\text{th}$ roots of the complex number are spaced $\frac{2\pi }{n}$ radians apart on a circle with radius $\sqrt[n]{r}$ .