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I need to find all complex numbers which satisfy the equation and I don't know how.

$$z^3 + 3z^2 + 3z = \overline z$$

Thanks for help

Maruša J
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1 Answers1

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Hint: write it as $z^3 + 3z^2 + 3z + 1= \overline {z + 1}$.

Hint #2: let $w=z+1$ then the equation becomes $w^3 = \overline w$.

Hint #3: $\;\;w^3 = \overline w \;\implies\; | w^3 | = | \overline w | \;\implies\; | w|^3 = | w |\;$ which means $|w|=0$ or $|w|=1$.

  • If $|w|=0$ then $w=0$ and $z=w-1=-1$ which is one solution to the original equation.

  • If $|w|=1 \iff w \overline w = 1$ then $\overline w = \frac{1}{w}$ and the equation becomes $w^4=1$, which gives $4$ more solutions in $w$ which in turn translate to $4$ more solutions in $z$.

dxiv
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  • Should I write: z as a+bi, because when I do so I come to a big equation. – Maruša J Oct 25 '16 at 19:25
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    $z^3+3z^2+3z+1=(z+1)^3$ – bigfocalchord Oct 25 '16 at 19:26
  • @MarušaJ You want to defer that $z=a+bi$ substitution for as long as possible, precisely because the equations become big. See my edit. – dxiv Oct 25 '16 at 19:27
  • I get bi=1 at the end. I don't know what I am doing wrong. – Maruša J Oct 25 '16 at 20:19
  • @MarušaJ Don't know since you didn't show the derivation, but maybe you turned to $z=a+bi$ too early. See the latest edit. – dxiv Oct 25 '16 at 20:35
  • @dxiv Uhh sorry but I'm kinda confused about this question as well , if $|w|=1$ then $|z+1|=1$ so that would be a circle of centre (-1,0) and radius 1? – bigfocalchord Oct 26 '16 at 02:45
  • @dydxx If you read to the end, $|w|=1$ is a necessary condition, but not a sufficient one. The necessary and sufficient condition is $w^4=1$. – dxiv Oct 26 '16 at 02:48