### Instruction: Find the solutions to $z^4 = -4.$  Enter the solutions, separated by commas.### Response: To find the solutions to this equation, I will use the fact that $z^4 = (re^{i\theta})^4 = r^4 e^{i4\theta}$, where $r$ and $\theta$ are the polar coordinates of $z$.
I will also use the fact that $-4 = 4 e^{i\pi}$, since the point $(-4,0)$ has a modulus of $4$ and an angle of $\pi$ radians.
Equating the two expressions, I get $r^4 e^{i4\theta} = 4 e^{i\pi}$.
This implies that $r^4 = 4$ and $4\theta = \pi + 2k\pi$, where $k$ is any integer.
Solving for $r$, I get $r = \sqrt[4]{4} = \sqrt{2}$.
Solving for $\theta$, I get $\theta = \frac{\pi}{4} + \frac{k\pi}{2}$.
To find the four distinct solutions, I will plug in different values of $k$.
For $k = 0$, I get $\theta = \frac{\pi}{4}$ and $z = \sqrt{2} e^{i\frac{\pi}{4}} = \sqrt{2} \left( \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i \right) = 1 + i$.
For $k = 1$, I get $\theta = \frac{3\pi}{4}$ and $z = \sqrt{2} e^{i\frac{3\pi}{4}} = \sqrt{2} \left( -\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i \right) = -1 + i$.
For $k = 2$, I get $\theta = \frac{5\pi}{4}$ and $z = \sqrt{2} e^{i\frac{5\pi}{4}} = \sqrt{2} \left( -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} i \right) = -1 - i$.
For $k = 3$, I get $\theta = \frac{7\pi}{4}$ and $z = \sqrt{2} e^{i\frac{7\pi}{4}} = \sqrt{2} \left( \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} i \right) = 1 - i$.
For any other value of $k$, the solution will be a repeat of one of the above, so I have found all four solutions.