Question

Given that $\zeta$ is the unit circle in counter-clockwise direction, evaluate:
\[ \oint_{\zeta} \frac{z^3}{4z - i} \, dz \] (round off to three decimal places).

Hint:

Always check whether the pole lies inside the contour before applying the residue theorem.

Answer 1

Correct Answer: 0.02

The integrand has a pole at:
\[ 4z - i = 0 $\Rightarrow$ z = \frac{i}{4}. \] Since \( \left|\frac{i}{4}\right| = 0.25 < 1 \), the pole lies inside the unit circle. Residue of \[ \frac{z^3}{4z - i} \] at \( z = \frac{i}{4} \) is:
\[ \text{Res} = \frac{z^3}{4} \Big|_{z = i/4} = \frac{1}{4} \left( \frac{i}{4} \right)^3 = \frac{1}{4} \cdot \frac{i^3}{64} = \frac{1}{4} \cdot \frac{-i}{64} = -\frac{i}{256}. \] By residue theorem:
\[ \oint = 2\pi i \left( -\frac{i}{256} \right) = 2\pi \left( \frac{1}{256} \right) = \frac{\pi}{128}. \] Numerically: \[ \frac{\pi}{128} = 0.02454 \approx 0.025. \]