Find the roots of $z^{3} =(3+4i)* \bar{z}$ So I thought that you can't write z with sin and cos, so I tried to write z as a+ib, but I got 2 relations and I don't know how I could factor or something. $a{^{3}}+3a^{2}ib-3ab^{2}-ib^{3}=3a-3ib+4ai+4b$ . It looks a bit off, I don't know, can someone help please?
 A: If $z^3=(3+4i)\overline z$, then$$\lvert z\rvert^3=\lvert z^3\rvert=\left\lvert(3+4i)\overline z\right\rvert=5\lvert z\rvert$$and therefore $\lvert z\rvert=0$ or $\lvert z\rvert^2=5$. In the first case, $z=0$, which is indeed a solution of the equation. Otherwise, $z=\sqrt 5\bigl(\cos(\theta)+\sin(\theta)i\bigr)$, for some $\theta\in\mathbb R$. Can you take it from here?
A: HINT
To begin with, notice that $z = 0$ is a solution. Let $z = re^{i\theta}$ and $r \neq 0$. Then $\overline{z} = re^{-i\theta}$. Since $3 + 4i = 5e^{i\varphi}$, we get the equation
\begin{align*}
r^{3}e^{3\theta i} = 5re^{-i\theta}e^{i\varphi} \Longleftrightarrow r^{2}e^{4i\theta} = 5e^{i\varphi} \Longleftrightarrow (r^{2} = 5)\,\wedge\,(4\theta = \varphi)
\end{align*}
Can you proceed from here?
A: As a rule of thumb, substituting $z=a+ib$ when solving an equation only makes things harder. Possible exceptions are when there are obvious simplifications to be done immediately, such as lots of $Re(z)$ and $Im(z)$ in the expression. But generally it is a thing best avoided.
In this case, it is probably better to use absolute values. You get
$$
|z^3|=5|z|
$$
which is to say, either $z=0$, or $|z|=\sqrt5$. Let's take a closer look at the second possibility:
$|z|=\sqrt5$ gives us $\bar z=\frac5z$, which we can insert into the equation. Thus it becomes
$$
z^3=\frac{5(3+4i)}z\\
z^4=15+20i
$$
which is solved through just taking fourth roots using your own favourite method.
