= Solution
{c}
The <Cauchy derivative formula> says that if $f$ is <holomorphic> on a <neighbourhood> of a closed <disk> and $z_0$ is inside its positively oriented boundary <circle> $\gamma$, then for $n\geq0$,
$$
\boxed{f^{(n)}(z_0)=\frac{n!}{2\pi i}\int_\gamma\frac{f(z)}{(z-z_0)^{n+1}}\,dz.}
$$
For $n=0$ this is the <Cauchy integral formula>. The same formula holds for a positively oriented simple closed <contour> when $f$ is <holomorphic> on a <neighbourhood> of the contour and its enclosed region.
In the present problem the <unit circle> and its closed interior lie inside $|z|<R$. Parametrize it by $z=e^{it}$, $0\leq t\leq2\pi$, so $dt=dz/(iz)$. For a positive integer $n$,
$$
\begin{aligned}
\int_0^{2\pi}f(e^{it})\cos(nt)\,dt
&=\frac1{2i}\int_{|z|=1}f(z)\bigl(z^{n-1}+z^{-n-1}\bigr)\,dz\\
&=\frac1{2i}\left(0+\frac{2\pi i}{n!}f^{(n)}(0)\right).
\end{aligned}
$$
The first <contour integral> is zero by the <Cauchy integral theorem>, since $f(z)z^{n-1}$ is <holomorphic> on the <disk>; the second is evaluated by the <Cauchy derivative formula>. Therefore
$$
\boxed{\int_0^{2\pi}f(e^{it})\cos(nt)\,dt=\frac{\pi}{n!}f^{(n)}(0),\qquad n\geq1.}
$$
Taking $n=1$ gives the requested $\pi f'(0)$.
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