<s>[INST] Solving $\overline{z}\cdot|z|\cdot z^5=8\sqrt{2}\left(-\sin\frac{\pi}{5}+i\cos\frac{\pi}{5}\right)^8$ I have this equation to solve:
$$\overline{z}\cdot|z|\cdot z^5=8\sqrt{2}\left(-\sin\frac{\pi}{5}+i\cos\frac{\pi}{5}\right)^8$$
Since $\overline{z}\cdot z = |z|^2$ and utilizing the de Moivre's formula this can be simplified to:
$$|z|^3z^4=8\sqrt{2}\left(\cos\frac{8\pi}{5}+i\sin\frac{8\pi}{5}\right)$$
$$|z|^7(\cos{4\alpha} + i\sin{4\alpha})=8\sqrt{2}\left(\cos\frac{8\pi}{5}+i\sin\frac{8\pi}{5}\right)$$
From here I thought just to compare $|z|^7 = 8\sqrt{2}$ and the sine, cosine part.

Is this the correct way to go about it or could it be done using some simpler method?

 [/INST] From the beginning:
Since $i^8 = 1$, we can multiply by $i^8$ to get $(- \cos(\pi/5) - i \sin (\pi/5))^8$ which is the same as $(\cos \pi/5 + i \sin \pi/5)^8$.
So now you have:
$$|z|^3z^4=8\sqrt{2} e^{8i \pi/5}$$
which means that $z$ has modulus $(8 \sqrt{2})^{1/7} = \sqrt{2}$. Now since $|z|^3$ is real, you have:
$$\arg z = \frac{1}{4} \arg z^4 = \frac{1}{4} \arg |z|^3z^4.$$
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