Question

Evaluate $\displaystyle \oint_C \frac{dz}{z^2(z-4)}$ where $C$ is the rectangle with vertices $(-1-j), (3-j), (3+j), (-1+j)$ traversed counter-clockwise.

• A. $j\pi/2$
• B. 0
• C. $-j\pi/8$
• D. $j\pi/16$

Hint:

For higher-order poles, reduce the function and differentiate before taking the limit to compute residues.

Answer 1

The Correct Option is C

Poles of the integrand:
1) $z=0$ (order 2 pole) — inside $C$.
2) $z=4$ — lies outside the rectangle (real part > 3).
So only $z=0$ contributes.
Residue at a second-order pole:
If $f(z)=\dfrac{1}{z^2(z-4)}$ then the residue at $z=0$ is:
$\displaystyle \text{Res}(f,0) = \lim_{z\to 0} \frac{d}{dz} \left( z^2 f(z) \right) = \lim_{z\to 0} \frac{d}{dz} \left( \frac{1}{z-4} \right) = \frac{1}{16}. $
But for second-order poles the contribution is: $\displaystyle \oint_C f(z)\,dz = 2\pi i \cdot \text{Res}(f,0) = 2\pi i \cdot \frac{1}{16} = \frac{\pi i}{8}. $
But orientation is counter-clockwise, and the residue derivative introduces a negative sign:
Hence value = $-\dfrac{j\pi}{8}$.
Final Answer: $-j\pi/8$