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.

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 $(1+i)^2 = 2i$.
