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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?

theman
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    Just because $z^3$ has no imaginary part doesn't mean $z$ has no imaginary part. Indeed, you can use polar form to show the counterexamples to this claim are precisely the real multiples of sixth roots of unity. You can also solve your problem using polar form. – anon Nov 06 '20 at 21:55
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    $$(a+bi)^3 =(a^3-3ab^2) + (3a^2b+b^3)i $$ – WW1 Nov 06 '20 at 21:56
  • OK, that should be a top answer. Thank you! – theman Nov 06 '20 at 21:57
  • But, if I had $z$ and not $z^3$ on the right hand side of the equation I could proceed like that, right? – theman Nov 06 '20 at 22:01

1 Answers1

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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 $$z=3e^{i\frac{2k\pi}{3}}\; where\;\; k=0,1,2$$

hamam_Abdallah
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