Solve $z^4 + 4 = 0$ I'm trying to solve it by using its polar form, but then I get
$$
\begin{align*}
z^4 &= (\rho e^{i\phi})^4 = \rho^4 e^{4i\phi}\\
    &= -4 = -4 e^{0i}\\
\end{align*}
$$
From the definition of equality of complex numbers, $\rho^4 = -4$ and $4\phi = 0 + 2\pi k$ for some $k \in \mathbb{Z}$. 
This would mean $\rho = \sqrt{-2}$ and $\phi = \pi k / 2$. I have no idea how to interpret that imaginary radius, and Wolfram Alpha says the angle is $\pi k / 4$. Should this equation be solved using this method? Am I missing some step, or are my calculations incorrect?
I've already read Solve $z^4+1=0$ algebraically, but I want to know whether this equation is solvable using this method or another method should be used.
 A: The modulus of a nonzero complex number is always positive. So from your equation you get
$$
\rho^4\,e^{4i\phi}=4(-1)=4e^{-i\pi}.
$$
From there you get $\rho^4=4$ and $4\phi=2k\pi-\pi$ for $k$ integer. So
$$
\rho=4^{1/4}=\sqrt2,\ \ \phi=\frac{k\pi}2-\frac\pi4=\frac{(2k-1)\pi}4,\ k=1,2,3,4.
$$
(adding multiplies of $4$ to $k$ yields multiples of $2\pi$ that will not affect $\phi$)
A: You want to find the fourth roots of $-4$, namely $z=(-4)^{\frac{1}{4}}$.
Here, $\theta=\pi$ and $r=|z|=4$. 
Hence, by De'Moivre we have $z=\sqrt{2}(\cos(\frac{\pi+2\pi k}{4})+i\sin(\frac{\pi+2\pi k}{4}))$ for $k=0,1,2,3$.
In polar we have, $z=e^{\frac{(\pi+2\pi k)}{4}i}$ for $k=0,1,2,3$.
A: Something that we all never learned in high school, but should have, is the amazing factorization $X^4+4=(X^2+2X+2)(X^2-2X+2)$. With this, the total factorization is easy.
A: Using that $x^2-y^2 = (x+y)(x-y)$ and $x^2+y^2 = (x+iy)(x-iy)$, we get
\begin{align}
z^4+4 &= (z^2+2i)(z^2-2i)\\
&= (z+i\sqrt{2i})(z-i\sqrt{2i})(z+\sqrt{2i})(z-\sqrt{2i})\\
&= (z-1+i)(z+1-i)(z+1+i)(z-1-i)
\end{align}
where the last equality is due to $(1+i)^2 = 2i$.
