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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?

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