# If $z\overline{z} = z^2$ Does it imply $z$ is real?

I'm particulary interested in this since I'm stuck on a proof of $AA^*=A^2 \Rightarrow A$ is Hermitian.

After applying Schur's Theorem.

I reached the point where upper triangular $$T = \begin{bmatrix}a &b & c\\ 0 &d & e\\ 0&0 & f\end{bmatrix}$$ also has the property $$TT^*=T^2$$ Comparing diagonal entries lead me to $$|b|^2+|c|^2 = 0 \Rightarrow b=c=0$$, and $$|e|^2=0 \Rightarrow e=0$$

Hence T is diagonal with entries $a,d,f$

All I need to do is prove $T$ is Hermitian and easily prove that $A$ is also Hermitian.

Which leads me to $$|a|^2=a\overline{a}=a^2$$$$|d|^2=d\overline{d}=d^2$$$$|f|^2=f\overline{f}=f^2$$

Can I conlude that they are real and hence $T^*=T^T=T$ ?

• Hint: If $z=0$, then you are done. Otherwise divide both sided by $z$. – Cheerful Parsnip Dec 8 '17 at 5:07
• $z^2 = z \bar z = |z|^2 \ge 0 \implies z = \pm |z| \in \mathbb{R}$ – dxiv Dec 8 '17 at 5:11

If $z=0$ then it's obviously real.

If $z\ne0$ then you can cancel it, so $\overline z=z$, so $z$ is still real.

• Thank you, just realized that I could've taken case by case as well! – bolt997 Dec 8 '17 at 5:44

Note that

$$z \overline z=|z|^2$$

Thus

$$z \overline z=z^2\implies z^2=|z|^2 \implies Im(z)=0$$

Letting $z=a+bi$ for $a,b\in \mathbb{R}$, then your the condition your question imposes is $$a^2+b^2=a^2-b^2+2abi\implies b^2=abi$$ which requires $b=0$ as if not, we have $$b=ai\not\in\mathbb{R}$$

Any complex number $z$ can be written as $re^{i\theta}$ where $r=|z|$ and $\theta=arg(z)$. Then what you have got is $r^2=r^2e^{i2\theta}$, which implies $e^{2i\theta}=1$ or $2\theta=2k\pi$ for some $k\in\Bbb{Z}$