Evaluate
$$\int _C\frac{ e^{iz}\, dz}{(z+2i)^7}$$
where C is given by the square with vertices $(\pm3,\ 3)$, $(\pm3, -3)$
I know that there is one singularity at $z=-2i$, which lies inside the contour $C$.
So, by Cauchy's Integral Theorem, we have: $$\int _C\frac{ e^{iz}\, dz}{(z+2i)^7}$$ $$=\int _C\frac{e^{iz\,}dz}{(z+2i)^{(6+1)}}$$ $$=\frac{2\pi i}{6!} f^{(6)}(-2i)$$ where $f(z)=e^{iz}$
Evaluating the above at $z=-2$, we obtain: $$\frac{2\pi i}{720} (-e^{-2i^2})$$ $$=\frac{\pi i}{360}(-e^2)$$
Is this right? I'm not sure if $f(z)$ does in fact equal $e^{iz}$ in this problem.
Any help would be greatly appreciated. Thank you!