Help with complex numbers $z^4 = i \bar z ^3$ I'm having trouble with complex numbers.
For example, I need to find all the solutions to 
$$ z^4 = i \bar z ^3$$
My attempt was
$$z= |z|\ \left(\cos(\alpha)+i\ \sin(\alpha)\right) $$
$$ z^4 = i \bar z ^3$$
$$z = i \left({\frac {\bar z}{z}}\right) ^3 $$
$$z = i \left({\frac {\bar z}{|z|}}\right) ^6 = i\left(\cos(-6\ \alpha)+i\ \sin(-6\ \alpha)\right) = i\ \cos(-6\ \alpha)- \sin(-6\ \alpha)$$
$$|z|\ \left(\cos(\alpha)+i\ \sin(\alpha)\right) = i\ \cos(-6\ \alpha)- \sin(-6\ \alpha)$$
From where you get
$$|z|\ \cos(\alpha)= - \sin(-6\ \alpha)$$
$$ |z| \sin(\alpha)= \cos(-6\ \alpha)$$
From that, I got:
$$\cos (7\alpha) = 0$$
Then
$$ 7\alpha= \frac {\pi}{2} + k\pi \qquad k\in  \mathrm {Z} \\
\alpha= \frac {\pi}{14} + \frac{\pi}{7}k$$
With this I get $|z| =1$
Is it okay?
And if it is, isn't there any easier way to do it? I think I'm overcomplicating things...
 A: You got to almost exactly the right place, It is easier  if we hold off introducing sines and cosines. 
There is the obvious solution $z=0$. We now look for the non-zero solutions.
For non-zero $z$, taking norms, we find that the norm of $z$ must be $1$.
If we multiply both sides by $z^3$, we get the equivalent equation $z^7=i$. Now it's essentially over. For $i=\cos \frac{\pi}{2}+i\sin \frac{\pi}{2}$. Use De Moivre's Formula to write down the $7$-th roots of $i$. 
A: When you start to involve powers of complex numbers, working with the exponential form tends to be easier:
$$z=|z|e^{i\phi}$$
The equation is:
$$|z|^4e^{4i\phi}=i|z|^3e^{-3i\phi}$$
From here, we have the soolution $z=0$, if $z\neq 0$ then we can divide by the module of $z$, and changing $i=e^{i\pi/2}$, we have:
$$|z|e^{7i\phi}=e^{i\frac{\pi}{2}+2k\pi i}\qquad k\in\mathbb{Z}$$
We conclude that $|z|=1$ and
$$\phi=\frac{\pi}{14}+\frac{2k\pi}{7}$$
The solutions will start repeating when $\phi$ goes all the way around ($2k\pi/7=2\pi$), which means that you will get the seven different solutions for $k=0,1,2,3,4,5,6$.
A: Let $z=re^{i\theta}.$ Clearly, $z=0$ is solution so let us assume that $z\neq0.$ So, from the given equation,
\begin{align}
& &r^4e^{4i\theta}=ir^3e^{-3i\theta}\\
&\implies& re^{7i\theta}=i\\
&\implies & r\cos(7\theta)+ir\sin(7\theta)=i\\
&\implies & r\cos(7\theta)=0 \implies \cos(7\theta)=0\implies 7\theta=(2n+1)\pi/2\\
&\text{and}& r\sin(7\theta)=1 \implies r=1
\end{align}
So $z=e^{i\theta}; \theta=(2n+1)\pi/14.$
A: Note that $i=e^{i\pi/2}$ and write $z=re^{i\theta}$. Thus $z^4=i\bar z^3$ becomes $r^4e^{i4\theta}=r^3e^{i(\pi/2-3\theta)}$, from which $r=1$ and $4\theta=\pi/2-3\theta+2k\pi$ that yelds to $\theta=\frac17(\pi/2+2k\pi),\;\;k=0,\dots,6$.
