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

Step 1: Spot the poles.
The integrand \(\dfrac{ze^{-z}}{(z-1)(z-2)}\) has simple poles at \(z=1\) and \(z=2\).

Step 2: See which poles are inside the circle.
The curve is \(|z|=\tfrac32\). Since \(|1|=1<\tfrac32\) and \(|2|=2>\tfrac32\), only \(z=1\) sits inside.

Step 3: Use the residue theorem.
The integral equals \(2\pi i\) times the residue at \(z=1\).

Step 4: Compute the residue at \(z=1\).
For a simple pole, drop the \((z-1)\) factor and put \(z=1\): \[\text{Res}=\left.\frac{ze^{-z}}{z-2}\right|_{z=1}=\frac{1\cdot e^{-1}}{1-2}=\frac{e^{-1}}{-1}=-\frac1e.\]
Step 5: Multiply by \(2\pi i\).
\[\oint_C=2\pi i\left(-\frac1e\right)=-\frac{2\pi i}{e}.\]
Step 6: Match the option.
This is option 1.
\[ \boxed{-\dfrac{2\pi i}{e}} \]