Question

The residue of \(\dfrac{z}{(z-a)(z-b)}\) at infinity is:

• A. \(\dfrac{a}{b}\)
• B. \(-\dfrac{b}{a}\)
• C. \(1\)
• D. \(-1\)

Hint:

Residue at infinity equals minus the sum of finite-pole residues; here that sum is \(1\).

Answer 1

The Correct Option is D

Step 1: The residue at infinity equals the negative of the sum of the residues at all finite poles:
\[\text{Res}_{z=\infty} f = -\big(\text{Res}_{z=a} f + \text{Res}_{z=b} f\big)\]
Step 2: Compute the simple-pole residues of \(f(z) = \dfrac{z}{(z-a)(z-b)}\):
\[\text{Res}_{z=a} = \frac{a}{a-b}, \qquad \text{Res}_{z=b} = \frac{b}{b-a} = -\frac{b}{a-b}\]
Step 3: Add them:
\[\frac{a}{a-b} - \frac{b}{a-b} = \frac{a-b}{a-b} = 1\]
Step 4: Apply the residue-at-infinity rule:
\[\text{Res}_{z=\infty} = -(1) = -1\]
\[\boxed{\text{Res}_{z=\infty} = -1}\]