Determine if \(\lim _{z \rightarrow 1}(1-z) \tan \frac{\pi z}{2}\) exists or not. If the limit exists, then find its value.
Let \(t=1-z\):
\[
\begin{aligned}
& \Rightarrow z=1-t \\
& z \rightarrow 1 \\
& \Rightarrow t=0
\end{aligned}
\]
Now, we have:
\[
\begin{aligned}
& \lim _{t \rightarrow 0} t \tan \left(\frac{\pi}{2}(1-t)\right) \\
& = \lim _{t \rightarrow 0} t \tan \left(\frac{\pi}{2}-\frac{\pi}{2} t\right) \\
& = \lim _{t \rightarrow 0} t \cot \left(\frac{\pi}{2}t\right) \\
& = \lim _{t \rightarrow 0} \frac{t}{\tan \frac{\pi}{2} t}
\end{aligned}
\]
Now, multiply both the numerator and denominator by \(\frac{2}{\pi}\):
\[
= \lim _{t \rightarrow 0} \frac{2}{\pi}\left(\frac{\frac{\pi}{2} t}{\tan \frac{\pi}{2} t}\right) = \frac{2}{\pi}
\]
So, the limit \(\lim _{z \rightarrow 1}(1-z) \tan \frac{\pi z}{2}\) exists, and its value is \(\frac{2}{\pi}\).