Sachchidanand Prasad

04-04-2023

Problem: Let $\gamma $ be the positively oriented circle $\left\{ z\in \mathbb{C} :\vert z \vert =2 \right\} $. Evaluate \[ \frac{1}{2\pi \iota }\int_{\gamma }\frac{e^{\iota \pi z}}{(z-1)(z-3\iota )^2}\mathrm{d} z. \]

Solution: We will use the Cauchy integral formula to evaluate the given integral. Recall that

Suppose $\gamma $ is a simple closed curve oriented positively and the function $f(z)$ is analytic on a region containing $\gamma $ and its interior. Then for any $z_0$ inside $\gamma $ \begin{equation}\label{eq:04Apr2023-1} \int _{\gamma }\frac{f(z)}{z-z_0} = 2\pi \iota f\left( z_0 \right). \end{equation}

In the given problem, note that \[ f(z) = \frac{e^{\iota \pi z}}{(z-3\iota )^2} \] is analytic inside the circle $\vert z \vert =2$. Therefore, using the Cauchy integral formula we have \begin{align*} \frac{1}{2\pi \iota }\int_{\gamma }\frac{e^{\iota \pi z}}{(z-1)(z-3\iota )^2}\mathrm{d} z & = f(1) \\ & = \frac{e^{\iota \pi}}{(1-3\iota )^2} \\ & = \frac{-1}{1 - 9 - 6\iota } \\ & = \frac{1}{8+6\iota } . \end{align*} Therefore, \[ \textcolor{blue}{\boxed{ \frac{1}{2\pi \iota }\int_{\gamma }\frac{e^{\iota \pi z}}{(z-1)(z-3\iota )^2}\mathrm{d} z = \frac{1}{8+ 6\iota } }} \]