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.

From either equation first deduce an equation for the absolute value $a=|z|\in\Bbb R_{\geq0}$. For the first one gets $a^3=a$ or $a(a^2-1)=0$ with solutions $a\in\{0,1\}$. For the second one gets $a^{n-1}=a$ or $a(a^{n-2}-1)=0$; this again has solutions $a\in\{0,1\}$, unless $n=2$ in which case every $a\in\Bbb R_{\geq0}$ is a solution, and the solution $a=0$ disqualifies for $n\leq1$ because $a^{n-1}$ is then undefined.
So the second equation with $n=2$ is special; it reads $z=i\bar z$ which has infinitely many solutions $z\in\{\,r(1+\mathbf i)\mid r\in\Bbb R\,\}$. Also $a=0$ leads to $z=0$ which is a solution to both equations in all cases, except $n\leq1$.
With these cases/solutions out of the way, what remains is an equation for $z$ with $|z|=1$. For such numbers one has $\bar z=z^{-1}$ and using that we can get an algebraic equation (no using complex conjugation) in both cases. The first equation becomes $z^3=z^{-1}$ or $z^4=1$ with solutions $z\in\{1,\mathbf i,-1,-\mathbf i\}$. The second equation becomes $z^{n-1}=\mathbf iz^{-1}$ or $z^n=\mathbf i$, with $n$ solutions $z\in\{\,\exp_1(\frac{1+4k}{4n})\mid k=0,1,\ldots,n-1\,\}$ where $\exp_1:\Bbb R\to\Bbb C$ abbreviates the function $t\mapsto2\pi\mathbf it$ (the notation is my private invention and meant to suggest "exponential to the base $1=\exp(2\pi\mathbf i)$").
Summary of the number of solutions: $5$ for the first problem, and for the second problem according to the value of $n$ as follows: $n+1$ solutions when $n>2$, $~n$ solutions when $n\leq1$, and infinitely many solutions when$~n=2$.
