# Solving a complex number equation with both $z$ and its conjugate $\bar z$

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!

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

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.

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.

• Solving the above system I got the $4$ solutions: $z=-2\pm\sqrt{12}i$, $z=0$, or $z=4$.
– DKal
Mar 25, 2014 at 6:53
• I got the first two solutions, but how did you get z=0 or z=4? Mar 25, 2014 at 7:16
• Ops nevermind i got it thanks a lot! Mar 25, 2014 at 7:18
• Wops again, i actually don't get how you got those, explanation please! Mar 25, 2014 at 7:23
• 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 $a=4$.
– DKal
Mar 25, 2014 at 7:27