Solve complex equation $z^4=a^{16}$ Let $a$ be some complex number. I have to solve equation
$$z^4=a^{16}$$
One is tempted to "simplify" it to $z=a^4$, so it is the solutions. But somebody told me, there are more solutions than that. Is it true and why? It seems counterintuitive.
 A: Yes, there are more solutions than that. To see that concretely, suppose $a=1$. Then your original equation is
$$ z^4 = 1 $$
which has the solutions $\{1,-1,i,-i\}$, but if you rewrite it to $z=1$, you will have lost three of the solutions.
What you can do is rewrite $z^4=a^{16}$ to
$$ z=ka^4 \text{ for some }k\text{ with }k^4=1\text{, that is, }k\in\{1,-1,i,-i\} $$
which gives you four separate equations to solve as you plug the possible $k$s in.
A: Note that every $a \in \mathbb{C}$ can be written as $a = re^{i\varphi}$, hence your equation delivers
$$z^4 = r^{16} e^{16i\varphi}$$
The complex roots are given by
$$z^n =  re^{i\varphi} \implies z_k = r^{\frac{1}{n}}e^{i\left( \frac{\varphi}{n} + \frac{2\pi k}{n}\right)} \qquad k=0,\dots,n-1$$
So you have $n$ solutions for the $n^{\text{th}}$ complex root. In your case
\begin{align*}
z_0 &= r^4 e^{i\left( \frac{16\varphi}{4} + 0 \right)} = \left( r, \varphi \right) = \left( r^4, 4\varphi \right)\\
z_1 &= r^4 e^{i\left( \frac{16\varphi}{4} + \frac{2\pi}{4} \right)} = \left( r, \varphi \right) = \left( r^4, 4\varphi + \frac{\pi}{2} \right)\\
z_2 &= r^4 e^{i\left( \frac{16\varphi}{4} + \frac{4\pi}{4} \right)} = \left( r, \varphi \right) = \left( r^4, 4\varphi + \pi \right)\\
z_3 &= r^4 e^{i\left( \frac{16\varphi}{4} + \frac{6\pi}{4} \right)} = \left( r, \varphi \right) = \left( r^4, 4\varphi + \frac{3\pi}{2} \right)\\
\end{align*}
A: We see that $16/4 = 4$, so we can substitute $t = a^4$, then we can rewrite $$z^4 = a^{16} \Leftrightarrow z^4 = t^4$$ This has the solutions Henning described above, the four roots of unity: $\{1,i,-1,-i\} $ times $t$. Just to give some clarity as why it works.
