Cayley transform of matrices in $SU(n)$ are traceless. Let $SU(n)$ be the group consisting of those complex $n\times n$-matrices $A$such that $det(A)=1$ and $A\cdot A^* = 1,$ where $A^*$ is the conjugate transpose.  
I am trying to show that given $A \in SU(n)$ such that $det(1+A) \neq 0$ ($1$ being the identity matrix), we have that the trace of $(1-A)*(1+A)^{-1}$ is zero. How do I prove this fact?  
I tried to look at the situation when $A$ is diagonal, but this gave me nothing.  

This is not true. The Cayley transform is skew-Hermitian and so its trace always has a zero real part, but the imaginary part of the trace can be nonzero. E.g.
\begin{align}
A&=\operatorname{diag}\left(-i,\ e^{-i\pi/3},\ ie^{i\pi/3}\right)
=\operatorname{diag}\left(-i,\ \frac{1-i\sqrt{3}}2,\ \frac{-\sqrt{3}+i}2\right),\\
K=(I-A)(I+A)^{-1}
&=\operatorname{diag}\left(i,\ \frac{i}{\sqrt{3}},\ -(2+\sqrt{3})i\right),\\
\operatorname{trace}(K)&=-i\left(1+\frac{2}{\sqrt{3}}\right)\ne0.
\end{align}
