Solving $\overline{z}\cdot|z|\cdot z^5=8\sqrt{2}\left(-\sin\frac{\pi}{5}+i\cos\frac{\pi}{5}\right)^8$

Question

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?

Answer 1

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:

$$\arg z = \frac{1}{4} \arg z^4 = \frac{1}{4} \arg |z|^3z^4.$$

Answer 2

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 $\theta = \dfrac{2\pi}{5}$

Answer 3

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.