complex analysis - Given $f(z) = int_{gamma} frac{sin zeta + e^{izeta}}{(zeta-z)^{2} } dzeta$, find $f'(frac{pi}{4})$ if $gamma$ is the unit circle

Question: Assume that $\gamma$ is the positively oriented unit circle $|z| = 1$ in $C$. Let

$f(z) = \displaystyle \int_{\gamma} \frac{\sin \zeta + e^{i\zeta}}{(\zeta-z)^{2} } d\zeta$

Find $f'(\frac{\pi}{4})$.

Comments: This is a problem from an old exam in complex analysis. I have not found similar problems in my textbook so I would be grateful if someone could show me how to solve it (is it a "Dirichlet problem"?). I am studying for an exam and I would like to know how to deal with this problem type. All input appreciated.

Answer

Define $g(z)=\sin z+e^{iz}$. Then $g(z)$ is analytic in a neighborhood of $\{|z|\le 1\}$, so by Cauchy's integral formula:
$$g'(z)=\frac{2}{2\pi i}\int_\gamma\frac{\sin \zeta+e^{i\zeta}}{(\zeta-z)^2}\, d\zeta$$
From this, we see that the $f(z)$ in the problem statement is $\pi i g'(z)$. Now differentiate $g(z)=\sin z+e^{iz}$ twice to get an expression for $f'(z)$:
$$f'(z)=\pi i g''(z)=\pi i\left(-\sin z-e^{iz}\right)$$
$$f'\left(\frac{\pi}{4}\right)=\pi i g''\left(\frac{\pi}{4}\right)=\pi i \left(-\frac{1}{\sqrt{2}}-e^{i\pi/4}\right)=\frac{\pi-2i}{\sqrt{2}}$$