How to find all complex solutions for $z^2 - z^{-2} = 4i$

Question

I've tried using the methods we have been shown in lectures but they don't turn out an answer, could someone please help?

Answer 1

HINT

We have that for $z\neq 0$

$$z^2-\frac1{z^2}=4i \iff z^4-4iz^2-1=0$$

then use the quadratic formula.

Answer 2

$z^2 - z^{-2} = 4i$ is equivalent to $z^4 - 4iz^2 - 1 = 0$ and set $X=z^2$

Answer 3

Alt. hint:   write it as $\;\displaystyle z^2 - \frac{1}{z^2} \color{red}{-2i} = 4i \color{red}{-2i} \;\iff\; \left(z - \frac{i}{z}\right)^2 = 2i = (1+i)^2\,$.