How to solve such equations: $z^4 = -4$, $z \in \mathbb{C}$? I have the following task:

Compute in each case all $z\in\mathbb{C}$ such that $z^4 = -4$ and $z^3 = 5i$.

I do not know how to solve such eqautions fast. Do you have any idea of how to solve such equations?
And one further question:
The multiplication of two complex numbers is defined as
$$wz := (u+iv)(x+iy)=(ux-vy)+i(vx+uy)$$
But what if w or z would be defined in the following ways:
$$w = (u-iv)$$ or $$w = (-u+iv)$$ 
Would the formula change?
Thank you for helping me out guys
 A: For solving these equations, you can use exponential coordinates: write both $z$ as $\rho e^{i\theta}$ with $\rho\in(0,+\infty)$ and $\theta\in(-\pi,\pi]$ (treat $z=0$ as a special case, checking explicitly if it is a solution or not) and $w=\rho_w e^{i\theta_w}$. Then the equation $z^n=w$ rewrites as
$$\rho^ne^{in\theta}=\rho_we^{i\theta_w},$$
leading to the system of equations
$$\left\{\begin{array}{c}
\rho^n=\rho_w\\
n\theta=\theta_w\,[2\pi]
\end{array}\right.
\iff
\left\{\begin{array}{c}
\rho=\sqrt[n]{\rho_w}\\
\theta=\frac{\theta_w}{n}\,[\frac{2\pi}{n}]
\end{array}\right.,$$
where $\rho$ is now explicit and you still have to find the $\theta$ checking that.
For example, in your first case $z^4=-4=4e^{i\pi}$, you should find $\rho=\sqrt[4]{4}$ and $\theta=\frac{\pi}{4}\,[\frac{\pi}{2}]$, which leads to $\theta\in\{\frac{\pi}{4},\frac{3\pi}{4},\frac{-\pi}{4},\frac{-3\pi}{4}\}$ if you use the info $\theta\in(-\pi,\pi]$.
A: You have to keep in mind that taking roots are multi-valued in the complex plane. By the fundamental theorem of algebra every non-constant polynomial of degree $n$ has exactly $n$ in the complex plane when counted with multiplicity. This applies also for polynomials of the form $p(z)=z^n-c$ for some complex number $c\in\Bbb C^*$.
Of particular interest are the so-called roots of unity which are the roots of polynomials of the form $p(z)=z^n-1$. Solutions to the equation $z^n-1=0$ are called the $n^\text{th}$ roots of unity. Using polar coordinates we find that these are of the form $\zeta_n^k=\exp\left(k\frac{2\pi}n\right)$ for $k=0,...,n-1$. 
Now, given an equation of the form $z^n-c=0$ we shall write $c$ as $c=r\exp(i\theta)$ where $r=|c|$ and $\theta=\arg c\in[0,2\pi)$. Then we can conclude that $z=r^\frac1n\exp \left(i\frac\theta n\right)$. But wait; there is more. In fact, every $z_k=r^\frac1n\exp \left(i\frac\theta n\right)\zeta_n^k$, $k=0,\dots,n-1$, is a solution too! We can express the general solution(s) therefore in the following manner
$$z_k=r^\frac1n\exp\left(i\frac{\theta+2\pi k}n\right),\,k=0,\dots,n-1$$
Can you take it from here?
A: $$z^3=5i \implies z^4=5zi \implies -4=5zi \iff z=-4/5i=\frac{4}{5}i$$
But substituting $z$ back in the equations concludes there're no solutions.
