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))^{*}