Proove that U is unitary (Cayley transform)

[skrat]

Function $f(t)=\frac{t-i}{t+i}$ for $t\in \mathbb{R}$ maps real ax into complex circle. Show that for any hermitian operator $H$ operator $U:=(H-iI)(H+iI)^{-1}$ is unitary (where $H+iI$ is reversible)

If I understand correctly $U$ is unitary when $U=U^{T}$ right?

So I tried to show that $U$ is unitary like this (hopefully it is ok):
$U=(H-iI)(H+iI)^{-1}$ because H hermitian than $H^{*}=H$
$U=(H^{*}-iI)(H^{*}+iI)^{-1}$ but Identity does not change if I conjugate it and transpose it
$U=(H^{*}-iI^{*})(H^{*}+iI^{*})^{-1}=((H-iI)^{-1}(H+iI))^{*}$
$U=((H+iI)^{-1}(H-iI))^{T}$ so $U=U^{T}$
Right or wrong?

Than I have to calculate the inverse function $f^{-1}=(\zeta ))i\frac{1+\zeta }{1-\zeta }$ and show that $f^{-1}(U)$ is hermitian operator for any unitary operator $U$, where $I-U$ is reversible operator.

And here I have no idea how to start... :/

 

[skrat]

Ok first part is clearly wrong.

For unitary operator this has to be true: $UU^{*}=I or U=(U^{*})^{-1}$. After realizing that I had no problem proving that $U$ is unitary.

But I am still having problems with second part: If $f^{-1}$ is hermitian, than $f^{-1}(U)=(f^{-1}(U))^{*}$
$f^{-1}(U)=i\frac{I+U}{I-U}=i(I+U)(I-U^{-1})$ but $U$ is unitary, so $UU^{*}=I$
$f^{-1}(U)=i(I+(U^{*})^{-1})(I-U^{*})=i(I+U^{-1})^{*}(I-U)^{*}$

now I don't know how to get rid of $U^{-1}$ and that - that comes from $i^{*}$ in order to prove that $f^{-1}(U)=(f^{-1}(U))^{*}$