Find $\frac{zf^{'}(z)}{f(z)}$, where $-1 \leq \alpha \leq 0 $ and $0< v < 1$
Given: $f(z)= \frac{1}{\pi}(-\log (1-vz)+ \alpha \log(1-vz^{-1}))$
and $f^{'}(z)= \frac{1}{\pi}\left(\frac{v}{1-vz} + \frac{\alpha v}{z^{2}-vz}\right)$
$$\implies \frac{zf^{'}(z)}{f(z)} = \frac{z(\frac{1}{\pi}\left(\frac{v}{1-vz} + \frac{\alpha v}{z^{2}-vz}\right)}{\frac{1}{\pi}(-\log (1-vz)+ \alpha \log (1-vz^{-1}))} = \frac{z\left(\frac{v}{1-vz} + \frac{\alpha v}{z^{2}-vz}\right)}{(-\log (1-vz)+ \alpha \log(1-vz^{-1}))}$$
Let $z=e^{it}$ and consider $\zeta=e^{it}$ and $0 \leq t<2 \pi$.
$$\frac{zf^{'}(z)}{f(z)}\implies \frac{(\frac{v \alpha}{e^{it}(1-ve^{-it})}+ \frac{ve^{it}}{1-ve^{it}})}{(-\log (1-ve^{it})+ \alpha \log(1-v-e^{it}))}$$
Where $e^{it}= \cos t+i \sin t$ and $e^{-it}=\cos t-i\sin t$.
$$\frac{zf^{'}(z)}{f(z)} \implies \frac{(\frac{v \alpha}{\cos t+i \sin t(1-v \cos t-i \sin t)}+ \frac{v \cos t+i \sin t}{1-v \cos t+i \sin t})}{-\log (1-v \cos t+i \sin t)+ \alpha \log(1-v \cos t-i \sin t)}$$
Now we can compute for $\frac{zf^{'}(z)}{f(z)}$:
$$\frac{zf^{'}(z)}{f(z)} \implies \frac{(\frac{v \alpha}{ \cos t+i \sin t(1-v \cos t-i \sin t)}+ \frac{v \cos t+i \sin t}{1-v \cos t+i \sin t})}{-\log (1-v \cos t+i \sin t)+ \alpha \log (1-v \cos t-i \sin t)}$$
Hi there I've been working on this problem for a while and I have come to a halt. I've been stuck trying to figure out on how to combine the common denominator. I was wondering if anybody can assistance me on this problem. I want to thank you ahead of time for your cooperation.