\Rightarrow f_{\tan Y} \left( z \right) = \frac{1}
{{1 + z^2 }}\sum\limits_{k \in \mathbb{Z}} {f_Y \left( {k\pi + \arctan z} \right)} = \frac{{e^{ - \arctan z} }}
{{1 + z^2 }}\sum\limits_{k \in \mathbb{Z}} {\frac{{e^{ - k\pi } }}
{{\left( {1 + e^{ - \left( {k\pi + \arctan z} \right)} } \right)^2 }}}