<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] Hint:
Use the fact
$$x^2-a^2=(x-a)(x+a)$$
and
$$x^2+a^2=(x-ai)(x+ai)$$
so
$$z^4-(2+3i)^4=0$$
$$\left ( x^2-(2+3i)^2 \right )\left ( x^2+(2+3i)^2 \right )=0$$
$$\left ( x-(2+3i) \right )\left ( x+(2+3i) \right )\left ( x-(2+3i)i \right )\left ( x+(2+3i)i \right )=0$$
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