What is $z^4=-4$ using the exponential form method, and with the answers written as $z = x + yi$? I know how to algebraically solve equations like this and find square roots of complex numbers, but how would you solve equations like that with the $n$-th power of $z$ being greater than $2$? 
 A: $z^4=(|z| e^{it})^4 = |z|^4 e^{4it} = -4\\
|z| = (4)^{\frac14} = \sqrt 2$
$e^{4it} = -1 = e^{\pi i}\\
4t = \pi$
$z=\sqrt 2 e^{\frac {\pi i}{4}} = 1 + i$
This is just one of the 4 solutions...the others
$z=\sqrt 2 e^{\frac {\pi i}{4}+\frac {n\pi}{2}} = (1 - i), (-1-i), (-1-i)$
A: In exponential form, $z^4 = -4 = 4 \text{ cis } \pi$, where $\text{cis } \theta$ is shorthand for $e^{i\theta} = \cos \theta + i \sin \theta$.  Then $z_0 = \sqrt{2} \text{ cis } \frac{\pi}{4}$ (since $\left(\sqrt{2}\right)^4 = 4$), and $z_k$ = $\sqrt{2} \text{ cis } \frac{(1+2k)\pi}{4}$, for $k = 1, 2, 3$.
In rectangular form, let $z = a+bi$ with real $a$ and $b$.  Then $z^4 = a^4+4a^3bi-6a^2b^2-4ab^3i+b^4$, which gives
$$
4a^3b-4ab^3 = 0
$$
and
$$
a^4-6a^2b^2+b^4 = -4
$$
Well, $a \not= 0$ and $b \not= 0$, since either $a$ or $b$ being zero would force $z^4$ to be a non-negative real value, so we can divide the first equation by $4ab$ to get
$$
a^2-b^2 = 0
$$
or $a^2 = b^2$.  Then the second equation becomes
$$
-4a^4 = -4
$$
or $a^4 = 1$, and $a = \pm 1$, meaning $b = \pm 1$.
