$T:$ $\mathbb{C}$ $\to$ $\mathbb{C}$ by $T(z)=\lambda z + \mu \bar{z}$ for every $z \in \mathbb{C}$ Let $\lambda , \mu$ $\in$ $\mathbb{C}$. Define $T:$ $\mathbb{C}$ $\to$ $\mathbb{C}$ by $T(z)=\lambda z + \mu \bar{z}$ for every $z \in \mathbb{C}$.
Show that if  $\vert T(z) \vert = \vert z \vert$ then $\lambda\mu =0$
I'm doing the first thing that comes to mind:
$\vert \lambda z+\mu\bar{z}\vert ^2=\vert z \vert ^2$ So we have $(\lambda\mu+\mu \bar{z})(\overline \lambda \overline z + \bar{\mu}z)=z\bar{z}$
How can I conclude the proof? Thanks.
 A: Note that $|T(1)| = |\lambda + \mu| = 1$ and therefore $(\lambda+\mu)(\bar{\lambda} + \bar{\mu}) = |\lambda|^2 + |\mu|^2 + \mu\bar{\lambda} + \lambda\bar{\mu} = 1$. Also $1 = |T(i)| = |\lambda -\mu|$ and therefore $1 = (\lambda-\mu)(\bar{\lambda} - \bar{\mu}) = |\lambda|^2 + |\mu|^2 - \mu\bar{\lambda} - \lambda\bar{\mu}$. Combining these two equations it follows that $\mu\bar{\lambda} + \lambda\bar{\mu} = 0$. 
But if $z$ is a complex number such that $z + \bar{z} = 0$ then $z$ is of the form $k\cdot i$. Hence we can assume that $\lambda = r$ and $\mu = s\cdot i$ for real values $r$ and $s$, or $\lambda = r\cdot i$ and $\mu = s$.
Hence, $1 = |\lambda +\mu|^2 = \lambda^2 + \mu^2$. Using this on any $z\in\mathbb{C}$ yields 
$$|z|^2 = |T(z)|^2 = (\lambda z + \mu\bar{z})(\bar{\lambda}\bar{z} + \bar{\mu}z) = |\lambda|^2|z|^2 + |\mu|^2|z|^2 + \mu\bar{\lambda}\bar{z}^2 + \lambda\bar{\mu}z^2 = |z|^2 + \mu\bar{\lambda}\bar{z}^2 + \lambda\bar{\mu}z^2$$
And therefore
$$\mu\bar{\lambda}\bar{z}^2 + \lambda\bar{\mu}z^2 = 0$$ 
This, and $|T(z)| = |z|$, is true whenever exactly one of $\lambda =0$ and $\mu = 0$ is satisfied. So assume both are non-zero $\lambda = r$ and $\mu = i\cdot s$ as above.
Then,
$$i\bar{z}^2 - iz^2 =0 \Rightarrow {\bar z}^2 = z^2$$ 
which cannot hold for every $z$. Hence at least one of $\lambda$ and $\mu$ is equal to 0.
A: When $\lambda=0$ we are done. So assume $\lambda\ne0$. Then
$$T(e^{i\phi})=\lambda e^{i\phi}\left(1+{\mu\over\lambda}e^{-2i\phi}\right)$$
and therefore
$$\bigl|T(e^{i\phi})\bigr|=|\lambda|\>\left|1+{\mu\over\lambda}e^{-2i\phi}\right|\ .\tag{1}$$
When $|T(z)|=|z|$ for all $z\in{\mathbb C}$ then for $\phi\in{\mathbb R}$ the right hand side should be $\equiv1$. Now the points $1+{\mu\over\lambda}e^{-2i\phi}$ with $\phi\in{\mathbb R}$ are lying on a circle with center $1$ and radius $\left|{\mu\over\lambda}\right|$. When $\mu\ne0$ the absolute value of $1+{\mu\over\lambda}e^{-2i\phi}$ is not constant along this circle; therefore the right hand side of $(1)$ cannot be $\equiv1$ for $\phi\in{\mathbb R}$. It follows that necessarily $\mu=0$ in this case.
