# Solving $\lvert z \rvert z^2 = \sqrt{2}(1-i)\overline{z}$

I haven't been able to solve the complex equation $$\lvert z \rvert z^2 = \sqrt{2}(1-i)\overline{z},$$despite trying writing $z$ in different forms and using $z\overline{z}=\lvert z\rvert^2$. I'm missing something again, what is it?

• Hint: Write $z$ in polar coordinates. Commented Apr 14, 2016 at 17:28
• ... knowing that $\sqrt{2}(1-i)=2e^{-i \pi/4}$ (why that ?) Commented Apr 14, 2016 at 17:33

Write $z=re^{i\theta}$ we have $r^3e^{2i\theta}=2e^{-\frac{-\pi i}{4}}re^{-i\theta}$ so $r^2e^{3i\theta}=2e^{-\frac{-\pi i}{4}}$

Therefore $r=\sqrt{2}$ and $\theta=\frac{-\pi}{12}+\frac{2k\pi}{3}, k\in \mathbb{Z}$

and a trivial solution $z=0$ is missing when I canceled $r$.

• There's also the trivial $z=0$ solution... Commented Apr 14, 2016 at 17:44

Compute the modulus: $$|\,|z|z^2|=|\sqrt{2}(1-i)\bar{z}|$$ becomes $$|z|^3=2|z|$$ so either $z=0$ or $|z|=\sqrt{2}$. Of course $z=0$ is a solution.

Suppose $z\ne0$ and write $z=\sqrt{2}u$, where $|u|=1$. Then $\bar{z}=\sqrt{2}\bar{u}=\sqrt{2}u^{-1}$ and the equation becomes $$\sqrt{2}u^2=(1-i)u^{-1}$$ that is, $$u^3=\cos\left(-\frac{\pi}{4}\right)+i\sin\left(-\frac{\pi}{4}\right)$$

• I love this, thank you. Commented Apr 14, 2016 at 18:06
• @Richard It's not that different from the other proposed solution; in general I prefer avoiding $e^{i\pi t}$ if possible, if only for clearer typesetting. Please, upvote troooll's answer as soon as possible. Commented Apr 14, 2016 at 18:08
• Yes I know, but I took a liking for the way you found the modulus. Anyway, I was going to do it as soon as I have enough reputation. Commented Apr 14, 2016 at 18:15