Solve $ \begin{cases} z^{8}=|z|^{7} \\ z=\bar z e^{i\frac{\pi }{2}} \end{cases}$ someone can help me with the following system:
$$ \begin{cases}
z^{8}=|z|^{7}  \\ 
z=\bar z e^{i\frac{\pi }{2}}
 \end{cases}$$ 
I found the solution $z=0$
In fact from the second equation:
$$z=i\bar z$$
$$z-i\bar z=0$$
$$z(1-iz\bar z)=0$$
$(1-iz\bar z)=0$ and $z=0$
$$i|z|^{2}=1$$
$$z^8=iz\bar z$$
Now I'm stuck.
Thank you a lot!
 A: From the first equation you have that $|z|^8=|z|^7$ which implies that $z=0$  or $|z|=1$. Let's find the other solutions on the unit circle by letting $z=e^{i\theta}$.
We have that $\overline{z}=e^{-i\theta}$ and the two equations become
$$e^{i8\theta}=1\quad\mbox{and}\quad e^{i2\theta}=e^{i\pi/2}.$$
The first one is solved by $\theta=2\pi k/8=k\pi/4$, for $k=0,1,\dots,7$.
As regards the second one $e^{ik\pi/2}=e^{i\pi/2}$ implies that $k=1$ or $k=5$.
Hence the solutions are three: $0$, $e^{i\pi/4}=\frac{1+i}{\sqrt{2}}$, and $e^{i5\pi/4}=-\frac{1+i}{\sqrt{2}}$. 
P.S. Note that from the system of equations, if $z$ is a solution then $-z$ is a solution too.
A: Say $z=re^{i\theta}$ in the polar form.
Putting this value of $z$ in the 2 given equations, we get that
$$z^{8}=|z|^{7} \implies r^{8}e^{i\cdot 8\theta}=r^{7} \implies e^{i\cdot 8\theta}=\frac{1}{r}\tag1$$ $$z=\bar z e^{i\frac{\pi }{2}} \implies z^2=z\bar z e^{i\frac{\pi }{2}} \implies r^{2}e^{i\cdot 2\theta}=r^2e^{i\frac{\pi }{2}} \tag2$$
So $(2)$ means $\theta=n\pi+\frac{\pi}{4}$ and $(1)$ means $\cos 8\theta=\frac{1}{r}$ and $\sin 8\theta =0 $
Hence $\cos 8\cdot(n\pi+\frac{\pi}{4})=1=\frac{1}{r} \implies r=1$
So $z=1\cdot e^{i(n\pi+\frac{\pi}{4})}=(-1)^n\cdot\frac{1+i}{\sqrt2}$ and apart from this , we have the trivial $z=0$.
