Solving $\begin{cases} z_1z_2=10\operatorname{cis}(\frac{4\pi}5)\\\frac{z_1}{\overline{z_2}^2}=\frac2{25}\operatorname{cis}(\frac{\pi}5)\end{cases}$ 
Solve
  $$\begin{cases}z_1z_2=10(\cos\frac{4\pi}{5}+i\sin\frac{4\pi}{5})\\\frac{z_1}{\overline{z_2}^2}=\frac{2}{25}(\cos\frac{\pi}{5}+i\sin\frac{\pi}{5})\end{cases}$$
  over $\mathbb C$. The answer should be in trigonometric form. $z=a+ib$.

Let:
$$
z_1=r_1(\cos\theta_1+i\sin\theta_1)\\
z_2=r_2(\cos\theta_2+i\sin\theta_2)\\
\frac{z_1}{\overline{z_2}^2}=\frac{z_1z_2^2}{\overline{z_2}^2z_2^2}=\frac{r_1r_2^2}{|z_2|^4}\stackrel{(\ast)}{=}\frac{r_1r_2^2}{r_2^4}=\frac{r_1}{r_2^2}
$$
then:
$$
\begin{cases}r_1r_2(\cos(\theta_1+\theta_2)+i\sin(\theta_1+\theta_2))=10(\cos\frac{4\pi}{5}+i\sin\frac{4\pi}{5})\\[10pt]
\dfrac{r_1(\cos\theta_1+i\sin\theta_1)}{r_2^2(\cos2\theta_2+i\sin2\theta_2)}\stackrel{(\ast\ast)}{=}\frac{r_1}{r_2^2}(\cos(\theta_1+2\theta_2)+i\sin(\theta_1+2\theta_2))=\frac{2}{25}(\cos\frac{\pi}{5}+i\sin\frac{\pi}{5})\end{cases}
$$
According to De Moivre's rule we know that $z^n=r^n \operatorname{cis}(n\theta)$ but  in the transition $(\ast)$ why does $|z_2|^4=r_2^4$?
Lastly in the transition $(\ast\ast)$ is there some trig identity which leads to that?
Please explain the points $(\ast)$ and $(\ast\ast)$.
 A: The line where you have $(*)$ should better be
$$
\left|\frac{z_1}{\overline{z_2}^{\,2}}\right|=
\frac{|z_1|}{\bigl|\overline{z_2}^{\,2}\bigr|}=\frac{r_1}{r_2^2}
$$
There is no reason for the equality before the one you mark with $(*)$.
The equality you mark with $({*}{*})$ is indeed wrong and it should be
$$
\frac{r_1(\cos\theta_1+i\sin\theta_1)}{r_2^2(\cos2\theta_2-i\sin2\theta_2)}=\frac{r_1}{r_2^2}(\cos(\theta_1+2\theta_2)+i\sin(\theta_1+2\theta_2))=\frac{2}{25}(\cos\frac{\pi}{5}+i\sin\frac{\pi}{5})
$$
because if $z=r(\cos\theta+i\sin\theta)$, then $\bar{z}=r(\cos\theta-i\sin\theta)$. Next apply the standard rules
\begin{gather}
\cos\theta-i\sin\theta=(\cos\theta+i\sin\theta)^{-1}\\[4px]
(\cos\alpha+i\sin\alpha)(\cos\beta+i\sin\beta)=
\cos(\alpha+\beta)+i\sin(\alpha+\beta)
\end{gather}

Further notes.
The equality
$$
\frac{z_1z_2^2}{\overline{z_2}^{\,2}z_2^2}=\frac{r_1r_2^2}{|z_2|^4}
$$
is generally false, because there's no reason for $z_1z_2^2$ to be real.
When you expand $\dfrac{z_1}{\overline{z_2}^{\,2}}$, you should write
$$
\frac{r_1(\cos\theta_1+i\sin\theta_1)}
{\bigl(\overline{r_2(\cos\theta_2+i\sin\theta_2}\bigr)^2}
=
\frac{r_1(\cos\theta_1+i\sin\theta_1)}
{(r_2(\cos\theta_2-i\sin\theta_2)^2}
=
\frac{r_1(\cos\theta_1+i\sin\theta_1)}
{r_2^2(\cos2\theta_2-i\sin2\theta_2)}
$$
A: $r_1r_2=10$ and $\frac{r_1}{r_2^2}=\frac{2}{25}$, which gives $r_1=2$ and $r_2=5$.
Let $z_1=2cis\theta_1$ and $z_2=5cis\theta_2$.
Thus, $\theta_1+\theta_2=\frac{4\pi}{5}+2\pi k$ and $\theta_1+2\theta_2=\frac{\pi}{5}+2\pi m$, where $\{k,m\}\subset\mathbb Z$.
Finally we obtain: $\theta_1=\theta_2=\frac{7\pi}{5}$
Your $*$ is true by the De Moivre's rule.
Your $**$ is true because $\frac{cis\theta}{cis\phi}=cis(\theta-\phi)$ and $\overline{cis\theta}=cis(-\theta)$.
$\frac{cis\theta}{cis\phi}=cis(\theta-\phi)$ because
$$\frac{cis\theta}{cis\phi}=\frac{cis\theta\overline{cis\phi}}{cis\phi\overline{cis\phi}}=cis(\theta-\phi)=$$ 
