Let $T \colon \mathbb{C} \to \mathbb{C}$ be an $\mathbb{R}$-linear map (where $\mathbb{C}$ is identified with $\mathbb{R}^2$ as usual).
(a) Show that there exist complex numbers $\lambda, \mu \in \mathbb{C}$ such that for any $z \in \mathbb{C}$, one has $$ T(z) = \lambda z + \mu \overline{z} \qquad \text{in $\mathbb{C}$}. $$ Also show that $\lambda$ and $\mu$ are uniquely determined by $T$, by giving explicit expressions of $\lambda$ and $\mu$ in terms of $T(1)$ and $T(i)$.
(b) Show that $T$ is an isometry (i.e. $|T(z)| = |z|$ for any $z \in \mathbb{C}$) if and only if $\lambda \mu = 0$ and $|\lambda + \mu| = 1$.
(Original text here.)
For part (a) I have shown that $\lambda = 0.5 (T(1)+iT(i))$ and $\mu = 0.5 (T(1)-iT(i))$.
For part (b) I just can't seem to show that $\lambda\mu=0$. I know that $$ |\lambda+\mu| = |T(1)| = |1|. $$
Suggestions?
Maybe it's possible to do it simpler, but it could go like this: We have \begin{align*} |\lambda z + \mu\overline z|^2 &= |\lambda|^2|z|^2 + 2\Re(\lambda\overline\mu z^2) + |\mu|^2|z|^2\\ &= |\lambda+\mu|^2|z|^2 + 2\Re(\lambda\overline\mu(z^2-|z|^2)). \end{align*} Now, it $|\lambda+\mu|=1$ and $\lambda\mu = 0$, we find that $|\lambda z + \mu\overline z|^2 = |z|^2$. Conversely, let $|\lambda z + \mu\overline z|^2 = |z|^2$ for all $z\in\Bbb C$. Then, letting $z=1$, we see that $|\lambda+\mu|=1$ and therefore also $\Re(\lambda\overline\mu(z^2-|z|^2)) = 0$ for all $z\in\Bbb C$. Setting $z=i$ and $z=i+1$ yields $\lambda\overline\mu = 0$ and thus $\lambda\mu = 0$.