$ |T z|= |z| \,\,\forall z\in \mathbb{C} \implies ab=0 $

Question

If $T: \mathbb{C} \to \mathbb{C}$ is given by $ T(z)= a z + b \bar z$ where $a,b \in \mathbb{C}$.

Then is it true that, $ |T z|= |z| \,\,\forall z\in \mathbb{C} \implies ab=0 $ ?

If yes, how can I show this? Any help would be appreciated. Thanks in advance.

Answer 1

Note \begin{align*} |Tz|^2 &= (Tz) \overline{T(z)}\\ &= (az+b \bar z) (\bar a \bar z + \bar b z)\\ &= |a|^2 z \bar z + |b|^2 z \bar z + a \bar b z^2 + \bar a b \bar z^2)\\ &= (|a|^2+|b|^2) |z| + (a \bar b z^2) + \overline{(a \bar b z^2)}\\ &= (|a|^2+|b|^2) |z| + 2 \text{Re}(a \bar b z^2).\\ \end{align*} If $a \bar b \neq 0$, then we can vary $z$ while keeping $|z|$ constant, and $|Tz|^2$ will change due to the $\text{Re($a \bar b z^2$)}$ term, but $|z|^2$ will not, a contradiction. So $a \bar b = 0$, giving either $a=0$ or $\bar b = 0$. The result follows.