For part a, since 0 is the singularity of the function, and it is inside the unit circle, why is the integration equal to $2\pi i$?
For part b, if we've proved part a, then we can substitute z with $cos\theta$ in it, but I get $2\pi$ instead of $\pi$. Thanks.
Hinzt: a) Follows from the Residue theorem.
b) Observe \begin{align} \int_C \frac{e^{az}}{z}\ dz =&\ \int^{\pi}_{-\pi} e^{a(\cos\theta+i\sin\theta)}e^{-i\theta}(ie^{i\theta})\ d\theta\\ =&\ \int^{\pi}_{-\pi} e^{a\cos\theta}[\cos a\sin \theta+ i\sin a\sin \theta]i\ d\theta= 2\pi i \end{align} which means \begin{align} 2\int^\pi_0e^{a\cos\theta} \cos a\sin \theta\ d\theta =\int^{\pi}_{-\pi}e^{a\cos\theta} \cos a\sin \theta\ d\theta = 2\pi. \end{align}