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 $z=r*e^{i\theta}$, So here we have,
$$\bar{z}*z^2*|z|=8\sqrt{2}(-sin\frac{\pi}{5}-icos\frac{\pi}{5})^8$$
$$(e^{-i\theta})(e^{i\theta})^2(r)=8\sqrt{2}(-i(cos\frac{\pi}{5}-isin\frac{\pi}{5}))^8$$
$$(e^{-i\theta})(e^{-\theta^2})(r)=8\sqrt{2}*1*(cos(-\frac{\pi}{5})+isin(-\frac{\pi}{5}))^8$$
$$(cos\theta-isin\theta)(e^{-\theta^2})(r)=8\sqrt{2}(e^{-i\frac{\pi}{5}})^8$$
$$(cos\theta)(e^{-\theta^2})(r)-(isin\theta)(e^{-\theta^2})(r)=8\sqrt{2}e^{-i\frac{8\pi}{5}}$$
$$(cos\theta)(e^{-\theta^2})(r)-(isin\theta)(e^{-\theta^2})(r)=8\sqrt{2}(cos(\frac{8\pi}{5})-isin(\frac{8\pi}{5}))$$
$$(cos\theta)(e^{-\theta^2})(r)-(isin\theta)(e^{-\theta^2})(r)=8\sqrt{2}cos(\frac{8\pi}{5})-i8\sqrt{2}sin(\frac{8\pi}{5})$$
So now we have segregated the real and complex parts,
So now, $(cos\theta)(e^{-\theta^2})(r)=8\sqrt{2}cos(\frac{8\pi}{5})$ and $(sin\theta)(e^{-\theta^2})(r)=8\sqrt{2}sin(\frac{8\pi}{5})$
On dividing Both Equations,
$$cot\theta=cot(\frac{8\pi}{5})$$
So from here we get $\theta=\frac{8\pi}{5}+2n$,where n is any integer.
So Now,
$$(cos(\frac{8\pi}{5}+2n))(e^{-\theta^2})(r)=8\sqrt{2}cos(\frac{8\pi}{5})$$
so both cosines will cancel out
$$(e^{-\theta^2})(r)=8\sqrt{2}$$
$$(r)=8\sqrt{2}(e^{\frac{8\pi}{5}^2})$$
Hence, $$z=8\sqrt{2}(e^{\frac{8\pi}{5}^2})(e^{i(\frac{8\pi}{5}+2n)})$$ where n is any positive integer.
