Complex numbers singularites

Question

I was recently watching a video about residue theorem, and I got confused on one part which he didn't explain in details

how does this $$z^4+1=0$$

lead to that $$z=e^{i(\frac{\pi}{4}+\frac{\pi}{2}n)}$$

where $n=0,1,2,3$

Answer 1

From $|z|=1$, let $z=\cos\theta+i\sin\theta$ then $z^4=\cos4\theta+i\sin4\theta=-1$ which shows $\cos4\theta=-1$ or $4\theta=2k\pi+\pi$ and then $$\theta=\dfrac{k\pi}{2}+\dfrac{\pi}{4}$$ so $$z=e^{i\left(\frac{k\pi}{2}+\frac{\pi}{4}\right)}$$ where $k=0,1,2,3$.