Find how many complex roots the equation has How many complex roots has each of the equations:
$$z^3 = \overline{z}$$
$$z^{n-1} = i \overline{z}$$
Where $\overline{z}$ is the complex conjugate of $z$.
For the first one I tried giving the form $z = a + bi,\:a, b \in \mathbb R$ and find the roots but I think there's something else because at the second equation I can't do that.

For the first equation, we have $$|z|^3=|\bar z|\implies |z|^3 = |z|\implies |z| \in\{0,1\}.$$
If $z\neq 0$, we can multiply the first equation by $z$ to get $z^4 = z\bar z = |z|^2 = 1$ and we have four different solutions $\{\pm 1,\pm i\}$. Since $z=0$ is solution of the original equation as well, this gives total of five solutions.
For the second equation, if $n = 1$, then $i\bar z = 1$ has unique solution.
If $n>2$, just like in the first equation, we conclude that $|z| \in\{0,1\}$. So, if $z\neq 0$, multiply the equation to get $z^n = i$ which has $n$ different complex roots. Adding $z = 0$ gives total of $n+1$ solutions.
The case $n=2$ is special, as Marc van Leeuwen kindly pointed out in the comments. The equation becomes $z = i\bar z$. In this case we can't conclude that $|z| \in\{0,1\}$. Let $z=x+iy$ for $x,y\in\Bbb R$. The equation becomes $$x+iy = i(x-iy)\iff x+iy = y + ix \iff x = y$$ and thus, there are infinitely many solutions and they are of the form $r(1+i),\ r\in\Bbb R$.
More generally, these are equations of the form $z^{n-1}=a\bar z$ for some complex $a$, $|a|=1$. For $n=1$ solution is unique, and for $n>2$, we have $|z|\in\{0,1\}$. Again, either $z=0$ or we can multiply the equation by $z$ to get $z^n = a$ which has precisely $n$ different complex solutions. The case $n=2$ is again a special case and needs to be treated separately:
$$x+iy = (\alpha+i\beta)(x-iy)\iff \begin{align} (\alpha-1)x+\beta y &= 0\\ \beta x - (\alpha+1)y &= 0\end{align}$$
