Find all solutions $z^2\overline{z}^3=32$ $z^2\overline{z}^3=32$
I have bought about the following way:


*

*to simplify it by $z^2\overline{z}^3=z^2\overline{z}^2\overline{z}=(z\overline{z})^2\overline{z}=|z|^4\overline{z}$

*to replace $z$ or by $z=x+iy$ or $z=re^{i\theta}$ which should I choose? I have tried both with no success 
 A: Yes we have that
$$z^2\overline{z}^3=z^2\overline{z}^2\overline{z}=(z\overline{z})^2\overline{z}=|z|^4\overline{z}=32$$
but then $\bar z=x$ must be real and since $x^5=32 \implies x=2$ the only solution is $z=2$.
A: Quite often it useful to see what can you say about an absolute value of a complex number and this is such a problem.
$$z^2\overline{z}^3=32\implies |z^2\overline{z}^3|=32\implies |z|^5 = 32\implies |z|=2$$
so $$z^2\overline{z}^3=32\implies |z|^4\overline{z}=32\implies \overline{z}=2\implies z=2$$
A: If you use $z=re^{i\theta}$ with $r \in \mathbb R_{\ge 0}$ and $\theta \in \mathbb R$
then $z^2\overline{z}^3=32$ gives you $r^5 e^{-i\theta}=32e^{i2n\pi}$ for integer $n$
so matching coefficients gives $r=2$ and $\theta = -2n\pi$ 
and thus solutions are of the form $z=2e^{-i2n\pi}$ and the only one is $z=2$ 
A: I like using
$z = re^{i\theta}; \tag 1$
we have
$\bar z = r e^{-i\theta}; \tag 2$
we are given that
$z^2 \bar z^3 = 32; \tag 3$
using (1) and (2) yields
$r^5 e^{-i\theta} = r^2 e^{2i \theta} r^3 e^{-3i \theta} = z^2 \bar z^3 = 32, \tag 4$
whence
$r^5 = \vert r \vert^5 = \vert r^5 e^{-i\theta} \vert = \vert 32 \vert = 32; \tag 5$
it follows that
$r = 2; \tag 6$
therefore, by (4),
$32 e^{-i\theta} = 32 \Longrightarrow e^{-i\theta} = 1 \Longrightarrow  e^{i\theta} = 1; \tag 7$
therefore (1) becomes
$z = 2. \tag 8$
A: We have
$$
z^2\bar z^3 = \left(z\bar z\right)^2\bar z = |z|^4\bar z = 32\Rightarrow \bar z = \alpha + 0(-j) = \alpha = z
$$
hence 
$$
\alpha^5 = 32\Rightarrow \alpha = 2 = z
$$
