# Evaluate $\oint_C f(z) \ dz$, where $C$ is the circle $\left|z - \dfrac{1}{2}\right| = \dfrac{3}{2}$ and $f(z) = \dfrac{5}{(2z + i)(z - 2i)}$.

I have the following problem:

Evaluate $\oint_C f(z) \ dz$, where $C$ is the circle $\left|z - \dfrac{1}{2}\right| = \dfrac{3}{2}$ and $f(z) = \dfrac{5}{(2z + i)(z - 2i)}$.

I would appreciate it if people could please check my solution.

My Solution

We have a pole at $z = 2i$ and a pole at $z = \dfrac{-i}{2}$. However, if we draw a diagram of $C$, then we can see that only the pole $z = \dfrac{-i}{2} \in C$.

$$f(z) = \dfrac{5}{(2z + i)(z - 2i)} = \dfrac{5}{2\left(z + \dfrac{i}{2}\right)(z - 2i)}$$

So our focus is the pole of order $1$ at $z = \dfrac{-i}{2}$.

Using Cauchy's Residue Theorem, we get the following:

\begin{align} \operatorname{Res}\left[f, \frac{-i}{2} \right] & = \frac{1}{(1 - 1)!} \lim_{z \to 1} \frac{d^0}{dz^0}\left[(z + i/2)^1 \frac{5}{2\left(z + \frac{i}{2}\right)(z - 2i)}\right] \\[10pt] & = \lim_{z \to 1} \dfrac{5}{2(z - 2i)} \\[10pt] & = \dfrac{5}{2 - 4i} \\[10pt] \therefore \oint_C f(z) \ dz & = 2\pi i \cdot \operatorname{Res}\left[f, \dfrac{-i}{2} \right] \\[10pt] & = 2\pi i \cdot \frac{5}{2 - 4i} \\[10pt] & = \pi i \cdot \frac{5}{1 - 2i} \end{align}

I would greatly appreciate it if people could please take the time to check my solution. If there are any errors, please specify what they are and what the correct solution is.

• The residue isn't correct.You shouldn't be looking at the limit near $1$, it should be near the pole whose residue you want to compute. Edit: You can check part of your answer here. – Git Gud Oct 8 '17 at 22:12
• @GitGud Thanks for the response. You're absolutely correct. I have made adjustments to my solution and now have $\oint_C f(z) \ dz = -2 \pi$. Is this correct? – The Pointer Oct 8 '17 at 22:17
• Yes, apart from the residue calculation everything else seems fine and the result is $-2\pi$. – Git Gud Oct 8 '17 at 22:56
• @GitGud Ok. Thank you very much for the assistance! – The Pointer Oct 8 '17 at 22:59
• You're welcome. – Git Gud Oct 8 '17 at 23:05