<s>[INST] Solving $z^4=(2+3i)^4$ To solve the equation, I calculated right side:
$z^4=(2+3i)^4=(-5+12i)^2=-119-120 i$
And then I get the correct answer:
$z_k=\underbrace{\sqrt[8]{119^2+120^2}}_{\sqrt{13}} \times Cis(\cfrac{\pi+\tan^{-1}(\frac{120}{119})}{4}+\cfrac{k \pi}{2}), k=0,1,2,3$
But, I am looking for a way to solve the equation $z^4=(2+3i)^4$ without expanding the right side. So I tried :
$z={ \left| r \right| }e^{i \theta}$
$r^4e^{4 \theta i}=(\sqrt{13} e^{(2k\pi+\tan ^{-1}(\frac{3}{2}))i})^4$
$r=\sqrt{13}$
$4\theta=4 \times {(2k\pi+\tan ^{-1}(\frac{3}{2}))}$
$\theta=2k\pi+\tan ^{-1}(\frac{3}{2})$
But I calculated the value of $\theta$ wrongly. How can I fix it?
 [/INST] Much simpler: $z^4 =(2+3i)^4= 1\cdot (2+3i)^4$
and $z = 1^{\frac 14} (2+3i)$, where $1^{\frac 14}$ is understood to mean the four complex fourth roots of $1$, namely $\pm 1, \pm i$.
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