Question

The value of the integral \(\displaystyle\oint_{C}\frac{z e^{-z}}{(z-1)(z-2)}\,dz\) where \(C:|z|=\dfrac{3}{2}\) is

• A. \(-\dfrac{2\pi i}{e}\)
• B. \(\dfrac{2\pi i}{e}\)
• C. \(2\pi i\,e\)
• D. \(-2\pi i\,e\)

Hint:

We use Cauchy's residue theorem. The integral around a closed curve equals \(2\pi i\) times the sum of residues at the poles that lie inside the curve. So first we check which poles are inside the circle.

Answer 1

The Correct Option is A

Concept:
We use Cauchy's residue theorem. The integral around a closed curve equals \(2\pi i\) times the sum of residues at the poles that lie inside the curve. So first we check which poles are inside the circle.

Step 1:
The integrand \(\dfrac{z e^{-z}}{(z-1)(z-2)}\) has simple poles at \(z=1\) and \(z=2\). The curve is the circle \(|z|=\dfrac{3}{2}=1.5\). The pole at \(z=1\) is inside (since \(1<1.5\)), but \(z=2\) is outside (since \(2>1.5\)). So only \(z=1\) counts.

Step 2:
Find the residue at \(z=1\). For a simple pole, multiply by \((z-1)\) and put \(z=1\): \[\text{Res}_{z=1} = \left.\frac{z e^{-z}}{z-2}\right|_{z=1} = \frac{1\cdot e^{-1}}{1-2} = \frac{e^{-1}}{-1} = -\frac{1}{e}.\]

Step 3:
Apply the residue theorem: \[\oint_C = 2\pi i \times \left(-\frac{1}{e}\right) = -\frac{2\pi i}{e}.\]

Answer: Option (1) — \(-\dfrac{2\pi i}{e}\).