Solution to a complex polynomial equation I am trying to find the solutions to the equation
$1 - \alpha * z^{-N} = 0$
The solutions are the zeros to a system that I am working on, I'm not sure how to solve this polynomial equation.  Any help would be much appreciated.  Thanks.
 A: First some trivial operations:
$$1 - \alpha * z^{-N} = 0$$
$$\alpha * z^{-N} = 1$$
$$z^{-N} = \frac{1}{\alpha}$$
$z^{-N}$ can be only zero if $z$ is zero, but $\frac{1}{\alpha}$ is never zero. So we can get the recipe of both sides:
$$z^N = \alpha$$
Which leads to
$$z=\sqrt[N]\alpha \cdot e^\frac{2\pi j k}{N} (k=0,...,N-1)$$
This last step might be a little bit complicated. The essence is that not only the N-th power of $\sqrt[N]\alpha$ is $\alpha$, but also if you rotate it on the complex plane, with zero to $N-1$ times, by $\frac{2\pi}{N}$ rad.
The most important thing to know about such basic signal processing tasks: it is not really new math. It is ordinary high school math with some custom notation and practices. There is no wizardry, understanding it from the high school level is not so hard, than learning, for example, the complex arithmetics. But you should learn the math.
A: $$\begin{align*}1-\alpha z^{-N} & =0\\
\\
z^N - \alpha &= 0 \\
\\
z^N &= \alpha = |\alpha|e^{i[\arg(\alpha)+2\pi n]}\\
\\
z &= |\alpha|^{\frac{1}{N}}e^{i\left[\frac{\arg(\alpha)}{N}+\frac{2\pi n}{N}\right]} \quad n \in \{0, \dots, N-1\} \\
\end{align*}$$
If $\alpha \in \mathbb{R}$ and $\alpha > 0$, then $\arg(\alpha) = 0$ and you get the slighty simpler expression
$$ z = \alpha^{\frac{1}{N}}e^{i\frac{2\pi n}{N}} \quad n \in \{0, \dots, N-1\}$$
