Find $\int_0^{\pi+2i} \cos(z/2) \; dz $?
What is the procedure for doing this problem?
I 'observed' that the derivative of $2\sin(z/2)$ is $\cos(z/2)$ so my answer was $2\sin(z/2)$ evaluated between $0$ and $\pi+2i$ which gives me $2\sin\left(\frac{\pi+2i}{2}\right)$
But wolfram alpha says the answer is $2\cos i\ldots$ So what am I doing wrong?