Complex Roots of $(z^2 - i)^2 = -4$

Question

I'm having a bit of trouble finding the complex roots of the equation

$$ (z^2 - i)^2 = -4 $$

I've first taken the root of both sides to come up with

$$ z^2 - i = -2i $$

and then attempted to isolate the $ z $ on the left side and I end up with the following options $$ z^2 = \sqrt{3} \\ z^2 = -1 \\ $$ I have them attempted to utizilize the formula

$$ r (cos(\frac{\alpha + 2\pi*k}{n}) + i*sin(\frac{\alpha + 2\pi*k}{n}))$$

where $ r = {\sqrt{3}, -1}$ and $\alpha = \frac{\pi}{2}$ based on the lack of a real part. Ending up with:

For $r = \sqrt{3}; \alpha = \frac{\pi}{2}; k = 0 $ $$ \sqrt{3}(cos(\frac{\pi}{4}) + i*sin(\frac{\pi}{4}) \\ = \frac{\sqrt{6}}{2} + i*\frac{\sqrt{6}}{2}\\ $$

For $r = \sqrt{3}; \alpha = \frac{\pi}{2}; k = 1 $ $$ \sqrt{3}(cos(\frac{5*\pi}{4}) + i*sin(\frac{5*\pi}{4}) \\ = -\frac{\sqrt{6}}{2} - i*\frac{\sqrt{6}}{2} \\ $$

For $r = -1; \alpha = \frac{\pi}{2}; k = 0 $

$$ -1(cos(\frac{\pi}{4}) + i*sin(\frac{\pi}{4}) \\ = -\frac{\sqrt{2}}{2} - i*\frac{\sqrt{2}}{2}\\ $$

For $r = -1; \alpha = \frac{\pi}{2}; k = 1 $

$$ -1(cos(\frac{5*\pi}{4}) + i*sin(\frac{5*\pi}{4}) \\ =\frac{\sqrt{2}}{2} + i*\frac{\sqrt{2}}{2}\\$$

However something is wrong here, as I should most certainly also get a root in the 2nd quadrant, which I seem to be unable to find for some reason.

Answer 1

Note that$$(z^2-i)^2=-4\iff z^2-i=\pm2i\iff z^2=3i\text{ or }z^2=-i.$$In the end, you will have four solutions (the square roots of $3i$ and of $-i$).

Answer 2

Don't lose track of signs, that's the key:

I've first taken the root of both sides to come up with

$$ z^2 - i = -2i $$

taking the roots only tells you that $z^2-i$ is either $-2i$ or $+2i$.

This yields two new equations, each of which will again have two solutions, hence four in total.

Answer 3

your equation can be factorized into $$- \left( {z}^{2}+i \right) \left( -{z}^{2}+3\,i \right) =0$$