Unitary mapping is biholomorphic

Question

If $\rho$ is unitary mapping (preserves Hermitian inner product on $\mathbb{C}^2$: $\langle \rho x, \rho y\rangle=\langle x,y \rangle$), how to prove that it is a biholomorphic mapping of the ball? I have to check that $\rho$ is bijective (which is, since $\rho^{-1}=\rho^*$), holomorphic and that $ \rho'(z)\neq 0$ but I don't know how? Any help is welcome. Thanks in advance.

Answer 1

Apparently we can assume the mentioned inner product is the standard inner product on $\mathbb{C}^2$ WLOG(this assumption is only for the sake of clarity). Now observe that $$\langle c\rho x+\rho y,\rho w \rangle = c\langle \rho x , \rho w \rangle + \langle\rho y ,\rho w \rangle=c\langle x , w \rangle + \langle y , w \rangle = \langle cx+y,w \rangle = \langle \rho(cx+y),\rho w \rangle$$ for arbitrary c, x, y and w, and note that $\rho$ is bijective, we have $$\langle （c\rho x+\rho y） - \rho(cx+y), （c\rho x+\rho y） - \rho(cx+y) \rangle = 0$$therefore $$c\rho x+\rho y = \rho(cx+y), $$hence $\rho$ is $\mathbb{C}$-linear. Combining with the fact that $||e^{i \theta_z}z +e^{i \theta_w} w|| = ||z+w||$ implies $e^{i (\theta_w - \theta_z)}w = w$ or $\bar w$ and that all the choices must be compatible(since $z,w$ are arbitrary), we've derived that $\rho$ is of the form of left-multiplication by a fixed unitary complex number $u$. It is clear now that $\rho$ is a biholomorphic mapping of the ball.