nth power and nth roots of a complex number
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Let $z_0 = x_0 + j y_0 = r_0 e^{j \theta_0}$. For any integer $n$, the $n$th power of $z$, $z^n$ is simply obtained by using the multiplication operation $n$ times. In the polar form, $z_0^n = r_0^n e^{j n \theta_0}$. Just like how the two real numbers $1$ and $-1$ have the same square, different complex numbers can have the same $n$th power.

Consider the set of distinct complex numbers $z_k = e^{j \theta_0 + \frac{2 \pi k}{n}}$. All the $z_k$s are different have the same $n$th power for $k=0,1,2,\ldots,n-1$. We can see this by raising $z_k$ to the $n$th power to get

\begin{align} z_k^n = \left(e^{j \theta_0 + \frac{2 \pi k}{n}}\right)^n = e^{j n \theta_0 + 2 \pi k} = e^{j n \theta_0} \end{align}

The $n$th root of $z$ is a bit more interesting and tricky. Any complex number $z$ which is the solution to the $n$th degree equation

$z^n - z_0 = 0$

is an $n$th root of $z_0$. The fundamental theorem of algebra states that an $n$th degree equation has exactly $n$ (possibly complex) roots. Hence, every complex number $z_0$ has exactly $n$, $n$th roots. These roots can be found by using the fact $e^{j \theta} = e^{j (\theta + 2 \pi k)}$.

\begin{eqnarray} \nonumber z^n = z_0 & \Rightarrow & r^n e^{jn\theta} = r_0 e^{j\theta_0} = r_0 e^{j(\theta_0 +2 k \pi)} \\ & \Rightarrow & r = \sqrt[n]{r_0}, \ \ , \theta = \frac{\theta_0+2k\pi}{n} \mbox{ for } k = 0,1,2,\ldots n-1 \end{eqnarray}

Clearly, computing $n$th roots is much easier in the polar form than in the cartesian form.


Find the third roots of unity $\sqrt[3]{1}$.

Since $1 = 1 e^{j0}$, this corresponds to $r_0 = 1, \theta_0 = 0$. Hence, the three roots of unity are given by

$r = 1, \ \ \theta = 0,\frac{2\pi}{3},\frac{4\pi}{3}$

In cartesian coordinates, they are $(1+j0)$, $\left(-\frac{1}{2}+j\frac{\sqrt{3}}{2}\right)$,$\left(-\frac{1}{2}-j\frac{\sqrt{3}}{2}\right)$. These are referred to as $1,\omega,\omega^2$ sometimes. The three roots are shown in the figure.


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