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.
 A: Hint:
First equation: either $z=0$, or you can rewrite the equation as
$$z^4=z\,\bar z=\lvert z\rvert^2$$
So use the exponential form $z=r\,\mathrm e^{i\theta}$ to obtain the equation
$$r^4\mathrm e^{4i\theta}=r^2.$$
Second equation: same method – either $z=0$ or it is equivalent to
$$z^n=i\lvert z\rvert^2\iff r^n\mathrm e^{in\theta}=r^2\mathrm e^{\tfrac{i\pi}2}.$$
A: 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:
$$x+iy = (\alpha+i\beta)(x-iy)\iff \begin{align} (\alpha-1)x+\beta y &= 0\\ \beta x - (\alpha+1)y &= 0\end{align}$$
A: 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$~n=2$.
