<s>[INST] qq, can I solve equations with complex numbers like that? So I have 2 examples:
$\frac{27|z|}{z}=\frac{1}{3}\bar z *z^3 \implies 27|z|=\frac{1}{3}\bar z *z^4 \implies 27|z|=\frac{1}{3} |z|^2*z^3 \implies \frac{27}{|z|}=\frac{1}{3} z^3 \implies \frac{27}{ \sqrt{a^2 + b^2}}=\frac{1}{3} (a+bi)^3 $
On the left hand side I see no imaginary part, therefore I know that: $ b = 0$ nad $ \frac{27}{ a}=\frac{1}{3} a^3 $
$8z(\bar z) = \bar z^5 \implies 8z|z|= \bar z ^5 \implies 8\frac{|z|^2}{\bar z}|z|= \bar z ^5 \implies 8|z|^3= \bar z ^6 \implies 8(\sqrt{a^2 + b^2})^3 = (a-bi)^6$
Same thing - so I know that: $ b = 0 $ , so:   $ 8a^3 = a^6$
Is that ok?
 [/INST] hint
Approach by polar expression with
$$z=re^{it}=r(\cos(t)+i\sin(t))$$
the equation becomes
$$27\frac{r}{re^{it}}=\frac 13.re^{-it}.r^3e^{3it}$$
or, after simplification
$$81=r^4e^{3it}=3^4e^{2ik\pi}$$
So, the solution of your equation is
$$z=3e^{i\frac{2k\pi}{3}}\; where\;\; k=0,1,2$$
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