Problem source

The function $\mathrm{f}$ is defined, for any complex number $z$, by 
\[
\mathrm{f}(z)=\frac{\mathrm{i}z-1}{\mathrm{i}z+1}.
\]
Suppose throughout that $x$ is a real number. 
\begin{questionparts}
\item Show that 
\[
\mathrm{Re}\,\mathrm{f}(x)=\frac{x^{2}-1}{x^{2}+1}\qquad\mbox{ and }\qquad\mathrm{Im}\,\mathrm{f}(x)=\frac{2x}{x^{2}+1}.
\]
\item Show that $\mathrm{f}(x)\mathrm{f}(x)^{*}=1,$ where $\mathrm{f}(x)^{*}$ is the complex conjugate of $\mathrm{f}(x)$. 
\item Find expressions for $\mathrm{Re}\,\mathrm{f}(\mathrm{f}(x))$ and $\mathrm{Im}\,\mathrm{f}(\mathrm{f}(x)).$ 
\item Find $\mathrm{f}(\mathrm{f}(\mathrm{f}(x))).$ 
\end{questionparts}

Solution source

\begin{questionparts}
\item \begin{align*}
&& f(x) &= \frac{ix-1}{ix+1} \\
&&&= \frac{ix-1}{ix+1} \frac{1-ix}{1-ix} \\
&&&= \frac{ix-1+x^2+ix}{1^2+x^2} \\
&&&= \frac{x^2-1}{x^2+1} + i \frac{2x}{x^2+1} \\
\Rightarrow && \textrm{Re}(f(x)) &=  \frac{x^2-1}{x^2+1} \\
&& \textrm{Im}(f(x)) &= \frac{2x}{x^2+1}
\end{align*}

\item \begin{align*}
&& f(x)f(x)^* &= \frac{ix-1}{ix+1} \frac{(ix-1)^*}{(ix+1)^*} \\
&&&= \frac{ix-1}{ix+1} \frac{-ix-1}{-ix+1} \\
&&&=  \frac{ix-1}{ix+1} \frac{-(ix+1)}{-(ix-1)} \\
&&&= 1

\item \begin{align*}
&& \begin{pmatrix} i & -1 \\ i & 1 \end{pmatrix}^2 &= \begin{pmatrix} -1-i & -i-1 \\ -1+i & -i+1 \end{pmatrix} \\
\Rightarrow && f(f(z)) &= \frac{-(1+i)(z+1)}{(-1+i)(z-1)} \\
&&&= \frac{2i}{2} \frac{z+1}{z-1} \\
&&&= i \frac{z+1}{z-1} \\
\Rightarrow && \textrm{Re}(f(f(x))) &= 0 \\
&& \textrm{Im}(f(f(x))) &= \frac{x+1}{x-1}
\end{align*}

\item \begin{align*}
&& \begin{pmatrix} i & -1 \\ i & 1 \end{pmatrix} \begin{pmatrix} i & i \\ 1 & -1 \end{pmatrix} &= \begin{pmatrix} -2 & 0 \\ 0 & -2 \end{pmatrix} \\
\Rightarrow && f(f(f(z))) &= z
\end{align*}

\end{questionparts}