Problem: The value of the integral
\[
\int _0^{2\pi } \exp \left( e^{\iota \theta } - \iota \theta \right) \mathrm{d} \theta
\]
is equals to
-
$2\pi \iota $
-
$2\pi$
-
$\pi$
-
$\iota \pi$
Solution: We substitute
\[
e^{\iota \theta } = z \implies \iota e^{\iota \theta }\mathrm{d} \theta = \mathrm{d} z \implies \mathrm{d}\theta = \frac{\mathrm{d} z}{\iota e^{\iota \theta }} = \frac{\mathrm{d} z}{\iota z}
\]
Thus,
\begin{align*}
\int _0^{2\pi } \exp \left( e^{\iota \theta } - \iota \theta \right) \mathrm{d} \theta & = \int _0^{2\pi } \frac{e^{e^{\iota \theta }}}{e^{\iota \theta }}\mathrm{d} \theta \\
& = \int _{\mathbb{S} ^1} \frac{e^z}{z} \left( \frac{\mathrm{d} z}{\iota z} \right) \\
& = \frac{1}{\iota }\int _\mathbb{S} ^1 \frac{e^z}{z^2}\mathrm{d} z = \frac{1}{\iota } \left( 2\pi \iota e^0 \right) = 2\pi .
\end{align*}
Here in the last step we used the Cauchy integral formula for derivatives.
First recall the
Cauchy integral formula
Let $\gamma $ be a simple closed curve and the function $f$ is an analytic on a region containing $\gamma$ and its interior. Let us assume $\gamma $ is oriented counterclockwise. The for every $z_0$ inside $\gamma $ we have
\[
f(z_0) = \frac{1}{2\pi \iota}\int _\gamma \frac{f(z)}{z-z_0}.
\]
A generalization to this which is called
Cauchy integral formula for derivatives is given by
If $f(z)$ and $\gamma $ satisfies the same hypothesis as for Cauchy's integral formula, then for all $z_0$ inside $\gamma $ we have
\[
f^{(n)}\left( z_0 \right) = \frac{n!}{2\pi \iota } \int _\gamma \frac{f(z)}{(z-z_0)^{n+1}}\mathrm{d} z.
\]
Thus the correct answer will be
Option B.