Question

Given $z = x + iy$, $i = \sqrt{-1}$. $C$ is a circle of radius $2$ centred at the origin and traversed anticlockwise. The value of \[ \frac{1}{2\pi i} \int_C \frac{1}{(z-i)(z+4i)}\, dz \] is ____________ (round off to one decimal place).

Hint:

Only consider residues of poles lying inside the contour.

Answer 1

Correct Answer: 0.2

The poles of \[ f(z) = \frac{1}{(z-i)(z+4i)} \] are at: \[ z = i, \qquad z = -4i. \] Circle $|z| = 2$ includes $i$ but excludes $-4i$. Residue at $z=i$:
\[ \text{Res}_{z=i} f(z) = \frac{1}{i + 4i} = \frac{1}{5i} = -\frac{i}{5}. \] Thus,
\[ \frac{1}{2\pi i} \int_C f(z)\, dz = -\frac{i}{5}. \] Magnitude rounded to one decimal place:
\[ \boxed{0.2} \]