Prove complex identity of $\frac{1}{z}$ $\log(\cdot)$ and $\sqrt{\cdot} $ are any branches of the logarithm and square root, respectively. Show that if $z \notin \{0, \pm i\}$, then the following identity holds:
$$\frac 1z =\cot\left [\frac{1}{i}\log\left (\sqrt{\frac{1+iz}{1-iz}}\right ) \right ]$$

Whenever applicable 
\begin{align}
\cot \left [\frac 1i \ln\left ( \sqrt{\frac {1+iz}{1-iz}} \right )\right ] &= i \cdot\frac {\exp \left [\ln\left ( \sqrt{\frac {1+iz}{1-iz}} \right )\right ] + \exp \left [-\ln\left ( \sqrt{\frac {1+iz}{1-iz}} \right )\right ]}{\exp \left [\ln\left ( \sqrt{\frac {1+iz}{1-iz}} \right )\right ] - \exp \left [-\ln\left ( \sqrt{\frac {1+iz}{1-iz}} \right )\right ]} = \\
&= i \cdot \frac {\sqrt{\frac {1+iz}{1-iz}} + \sqrt{\frac {1-iz}{1+iz}}}{\sqrt{\frac {1+iz}{1-iz}} - \sqrt{\frac {1-iz}{1+iz}}} = i \cdot \frac {1+iz + 1 - iz}{1+iz - 1 + iz} = \\
&= i \cdot \frac 2{2iz} = \frac 1z
\end{align}
Some info
\begin{align}
\cot z &= \frac {\cos z}{\sin z} = \frac {\frac {e^{iz} + e^{-iz}}2}{\frac {e^{iz} - e^{-iz}}{2i}} = i \cdot \frac {e^{iz} + e^{-iz}}{e^{iz} - e^{-iz}} \tag 1 \\
-\ln z &= \ln \frac 1z \tag 2
\end{align}
