Back to Mth Roots

As mentioned in §3.4, there are $M$ different numbers $r$ which satisfy $r^M=a$ when $M$ is a positive integer. That is, the $M$ th root of $a$ , which is written as $a^{1/M}$ , is not unique--there are $M$ of them. How do we find them all? The answer is to consider complex numbers in polar form. By Euler's Identity, which we just proved, any number, real or complex, can be written in polar form as

$$z = r e^{j\theta}$$

where $r\geq 0$ and $\theta\in[-\pi,\pi)$ are real numbers. Since, by Euler's identity, $e^{j2\pi k}=\cos(2\pi k)+j\sin(2\pi k)=1$ for every integer $k$ , we also have

$$z = r e^{j\theta} e^{j2\pi k}.$$

Taking the $M$ th root gives

$$z^{\frac{1}{M}} = \left(r e^{j\theta} e^{j2\pi k}\right)^{\frac{1}{M}} = r^{\frac{1}{M}} e^{j\frac{\theta}{M}} e^{j2\pi \frac{k}{M}} = r^{\frac{1}{M}} e^{j\frac{\theta+2\pi k}{M}}, \quad k\in \mathbb{Z}.$$

There are $M$ different results obtainable using different values of $k$ , e.g., $k=0,1,2,\dots,M-1$ . When $k=M$ , we get the same thing as when $k=0$ . When $k=M+1$ , we get the same thing as when $k=1$ , and so on, so there are only $M$ distinct cases. Thus, we may define the $k$ th $M$ th-root of $z=r e^{j\theta}$ as

$$r^{\frac{1}{M}} e^{j\frac{\theta+2\pi k}{M}}, \quad k=0,1,2,\dots,M-1.$$

These are the $M$ $M$ th-roots of the complex number $z=r e^{j\theta}$ .