Evaluate the given integral along the indicated closed contour $\oint_C\frac{sinz}{(z^2)+(\pi)^2}dz; |z-2i|=2$

Question

Evaluate the given integral along the indicated closed contour $\oint_C\frac{sinz}{(z^2)+(\pi)^2}dz; |z-2i|=2$

This should be solve by using Cauchy's formula but i couldnt find.( z=$\pi$i is in C)

This is my solution way:

$\oint_C\frac{(sinz)/(z+(\pi)i)}{z-(\pi)i}dz$

If i say f(z)=$\frac{sinz}{z+(\pi)i}$

My answer will be $2\pi$i *f '($\pi$i) But i couldnt find f'($\pi$i)

Answer 1

Cauchy's integral formula is: $$f(a)=\frac{1}{2\pi i}\oint_C\frac{f(z)}{z-a}dz,$$ so taking $f(z)=\frac{\sin(z)}{z+\pi i}$: $$\frac{\sin(\pi i)}{2\pi i}=\frac{1}{2\pi i}\oint_C\frac{f(z)}{z-\pi i}dz=\frac{1}{2\pi i}\oint_C\frac{\sin(z)/(z+\pi i)}{z-\pi i}dz,$$ and hence your integral is equal to $\sin(\pi i)$ - no derivatives necessary.