Evaluate using cauchy's integral formula

Question

How can we evaluate this expression using cauchy's integral formula $\int_C \frac{e^{\pi Z}}{ ( {Z^2 + 1}) ^2} dZ$
where $C$ is $|Z-i|=1$

Answer 1

Clearly $$ \int_{|z-i|=1} \frac{e^{\pi z}}{ ( {z^2 + 1}) ^2} dz=\int_{|z-i|=1} \frac{\frac{\mathrm{e}^{\pi z}}{(z+i)^2}}{ ( {z-i}) ^2} dz. $$

According to the Cauchy Integral formula $f'(a)=\frac{1}{2\pi i}\int_{|z-a|=r}\frac{f(z)}{(z-a)^2}dz$ we have for $a=i$: \begin{align} \frac{1}{2\pi i}\int_{|z-i|=1} \frac{\frac{\mathrm{e}^{\pi z}}{(z+i)^2}}{ ( {z-i}) ^2} dz &=\left(\frac{\mathrm{e}^{\pi z}}{(z+i)^2}\right)'_{z=i}=\left(\frac{\pi\mathrm{e}^{\pi z}}{(z+i)^2}-2\frac{\mathrm{e}^{\pi z}}{(z+i)^3}\right)_{z=i} \\ &=\frac{\pi\mathrm{e}^{\pi i}}{(i+i)^2}-2\frac{\mathrm{e}^{\pi i}}{(i+i)^3}=\frac{-\pi}{-4}-2\frac{-1}{-8i} =\frac{\pi}{4}+\frac{i}{4}. \end{align} Thus $$ \int_{|z-i|=1} \frac{\frac{\mathrm{e}^{\pi z}}{(z+i)^2}}{ ( {z-i}) ^2} dz=\frac{\pi^2 i}{2}-\frac{\pi}{2}. $$