What is the Cauchy's integral formula for this?

Question

$$\oint_{c : |z -\pi i| = 1} \frac{e^{iz}}{z(z-\pi i)^2} dz $$

I tried to solve it and the answer is $-\frac{1}{\pi} + i$ How can I make sure that the answer is right ?

Answer 1

z = 0 is not inside the contour. That leaves only one pole inside the contour at $z = \pi i$

Cauchy integral formula says that is $f$ is holomorphic, for every $a$ inside the contour,

$\oint \frac {f(z)}{(z-a)^{n+1}} \ dz = \frac {2\pi i}{n!} f^{(n)}(a)$

$\frac {d}{dz} \frac {e^{iz}}{z} = \frac {e^{iz}(iz - 1)}{z^2}$

Evaluated at $z = \pi i$

$\frac {2\pi i}{1!}\frac {e^{-\pi}(-\pi - 1)}{-\pi^2}\\ \frac {2ie^{-\pi}(\pi + 1)}{\pi}$