1
$\begingroup$

Determine all possible values of $z\in\mathbb{C}$ that satisfy the equation $4z = \overline{z}^2$.

Where $\overline{z}$ represents the complex conjugate.

(Hint: There are $4$ solutions.)

Observations

If we had $4z=z^2$, that would be an easy quadratic equation, with solutions $0,4$.

And if it was $4\bar z = \bar z^2$, then after substitution $\zeta=\bar z$ we have a quadratic equation again.

But this equation has both $z$ and $\bar z$. I'm not sure how to solve these types of problems. Any tips or how to do these would be great thanks!

$\endgroup$
1
$\begingroup$

Denote $z=a+bi$ with $a,b\in\mathbb{R}$, then your equation says: $$4(a+bi)=(a-bi)^2.$$ Which means that $$4a+4bi=(a^2-b^2)-2abi.$$ Now the above equality of complex numbers is a system of equations of real numbers: $$\begin{cases} 4a=a^2-b^2\\ 4b=-2ab \end{cases}.$$ Solve it and you'll find the solutions.

$\endgroup$
  • $\begingroup$ Solving the above system I got the $4$ solutions: $z=-2\pm\sqrt{12}i$, $z=0$, or $z=4$. $\endgroup$ – DKal Mar 25 '14 at 6:53
  • $\begingroup$ I got the first two solutions, but how did you get z=0 or z=4? $\endgroup$ – user136088 Mar 25 '14 at 7:16
  • $\begingroup$ Ops nevermind i got it thanks a lot! $\endgroup$ – user136088 Mar 25 '14 at 7:18
  • $\begingroup$ Wops again, i actually don't get how you got those, explanation please! $\endgroup$ – user136088 Mar 25 '14 at 7:23
  • $\begingroup$ Consider the two cases: $b\neq 0$ or $b=0$. In the first case, the second equation implies that $a=-2$, substituting that into the first equation you get $b=\pm\sqrt{12}$. In the second case, $b=0$, the first equation implies $a^2-4a=0$ whence $a=0$ or $a=4$. $\endgroup$ – DKal Mar 25 '14 at 7:27
3
$\begingroup$

Hint: use the polar representation $z = r e^{i\theta}$.

$\endgroup$
2
$\begingroup$

There is the obvious solution $z=0$. From now on, assume $z\ne 0$. Taking norms, we find that $|z|=4$. Let $z=4e^{i\theta}$. Then we want $e^{i\theta}=e^{-2i\theta}$. We leave the rest to you.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.