Sachchidanand Prasad

14-02-2023

Problem: Let \[ f(z) = \frac{(z-2)^3 z^3}{(z+5)^3(z+1)^3(z-1)^4}. \] Compute the following integral. \[ \int_{\vert z \vert =3} \frac{f^\prime (z)}{f(z)}\mathrm{d} z. \]

Solution: We will use the Argument principle to solve this problem.

Let $\gamma $ be a simple closed curve, oriented counterclockwise direction. Suppose that $f$ is an analytic function on and inside $\gamma$ except for (possibly) some finite poles inside (not on ) $\gamma $ and some zeros inside (not on) $\gamma $. Then we have \[ \int_\gamma \frac{f^\prime (z)}{f(z)}\mathrm{d} z = 2\pi \iota \left( Z-P \right) , \] where $Z$ and $P$ denotes the number of zeros and poles of $f(z)$ inside $\gamma $ respectively, with each zero and pole counted with multiplicity.

Therefore, we have \begin{equation}\label{eq:14Feb2023-1} \int_{\vert z \vert =3} \frac{f^\prime (z)}{f(z)}\mathrm{d} z = 2\pi \iota \left( Z-P \right). \end{equation} Here we have \[ \textup{Zeros} = \{2,2,2,0,0,0\},~\textup{Poles} = \{-1,-1,-1,1,1,1,1\}. \]

Argument principle

Thus, $Z=6$ and $P=7$. Therefore, using \eqref{eq:14Feb2023-1} we will get, \[ \int_{\vert z \vert =3} \frac{f^\prime (z)}{f(z)}\mathrm{d} z = 2\pi \iota (6-7) = -2\pi \iota. \]