### Instruction: Find all solutions $x$ (real and otherwise) to the equation
\[x^4+64=0.\]Enter all the solutions, separated by commas.### Response: This problem is asking me to find the fourth roots of -64, which are complex numbers.
I recall that any complex number can be written in polar form as $r(\cos\theta+i\sin\theta)$, where $r$ is the modulus and $\theta$ is the argument of the complex number.
If I raise both sides of the polar form to the fourth power, I get \[r^4(\cos4\theta+i\sin4\theta)=-64.\]
This means that $r^4=64$ and $\cos4\theta+i\sin4\theta=-1$.
The first equation implies that $r=\sqrt[4]{64}=2\sqrt{2}$.
The second equation implies that $4\theta$ is an odd multiple of $\pi$, since that is when the cosine is -1 and the sine is 0.
So I can write $4\theta=(2k+1)\pi$, where $k$ is any integer.
Dividing by 4, I get $\theta=\frac{(2k+1)\pi}{4}$.
Now I can plug in different values of $k$ to get different solutions for $x$.
If $k=0$, I get $\theta=\frac{\pi}{4}$ and $x=2\sqrt{2}(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4})=2\sqrt{2}(\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2})=2+i2$.
If $k=1$, I get $\theta=\frac{3\pi}{4}$ and $x=2\sqrt{2}(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4})=2\sqrt{2}(-\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2})=-2+i2$.
If $k=2$, I get $\theta=\frac{5\pi}{4}$ and $x=2\sqrt{2}(\cos\frac{5\pi}{4}+i\sin\frac{5\pi}{4})=2\sqrt{2}(-\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2})=-2-i2$.
If $k=3$, I get $\theta=\frac{7\pi}{4}$ and $x=2\sqrt{2}(\cos\frac{7\pi}{4}+i\sin\frac{7\pi}{4})=2\sqrt{2}(\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2})=2-i2$.
These are the four solutions I was looking for, and they are all distinct.