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?


Let $z = A e^{i \theta} $ then
$ A^7 e^{i 4 \theta} = 8 \sqrt{2}(e^{-i \dfrac{3 \pi}{10}} )^8 = 2^{7/2}  e^{-i \dfrac{12 \pi}{5}}  = 2^{7/2}  e^{i \dfrac{8 \pi}{5}} $
Hence, $A = \sqrt{2}$, and $\theta = \dfrac{2\pi}{5}$
