### Instruction: Find the solutions to $z^4 = -4.$  Enter the solutions, separated by commas.### Response: I recognize that this is a complex equation, since the right-hand side is negative.
I also recall that any complex number can be written in polar form as $r e^{i \theta}$, where $r$ is the modulus and $\theta$ is the argument.
So I can rewrite the equation as $z^4 = 4 e^{i \pi}$, since $-4 = 4 e^{i \pi}$.
Taking the fourth root of both sides, I get $z = \sqrt{2} e^{i \frac{\pi}{4} + i \frac{k \pi}{2}}$, where $k$ is an integer.
This means that there are four possible values of $z$, depending on the value of $k$.
If $k = 0$, then $z = \sqrt{2} e^{i \frac{\pi}{4}} = \sqrt{2} (\cos \frac{\pi}{4} + i \sin \frac{\pi}{4}) = \frac{\sqrt{2}}{\sqrt{2}} + i \frac{\sqrt{2}}{\sqrt{2}} = 1 + i.$
If $k = 1$, then $z = \sqrt{2} e^{i \frac{3 \pi}{4}} = \sqrt{2} (\cos \frac{3 \pi}{4} + i \sin \frac{3 \pi}{4}) = \frac{-\sqrt{2}}{\sqrt{2}} + i \frac{\sqrt{2}}{\sqrt{2}} = -1 + i.$
If $k = 2$, then $z = \sqrt{2} e^{i \frac{5 \pi}{4}} = \sqrt{2} (\cos \frac{5 \pi}{4} + i \sin \frac{5 \pi}{4}) = \frac{-\sqrt{2}}{\sqrt{2}} + i \frac{-\sqrt{2}}{\sqrt{2}} = -1 - i.$
If $k = 3$, then $z = \sqrt{2} e^{i \frac{7 \pi}{4}} = \sqrt{2} (\cos \frac{7 \pi}{4} + i \sin \frac{7 \pi}{4}) = \frac{\sqrt{2}}{\sqrt{2}} + i \frac{-\sqrt{2}}{\sqrt{2}} = 1 - i.$
Therefore, the solutions are $z = 1 + i, -1 + i, -1 - i, 1 - i.$