GATE 2025 PH – Question 21
Consider $f(z)=1/[z^2(z-2)^3]$ of a complex variable $z$. The residues at $z=0$ and $z=2$, respectively, are
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Correct answer: (C) $-3/16,3/16$
Explanation
$f(z)=\dfrac1{z^2(z-2)^3}$ has a double pole at $z=0$ and a triple pole at $z=2$.
**Residue at $z=0$** (pole of order 2):
$$\text{Res}_{z=0}=\frac1{1!}\frac{d}{dz}\left[\frac{1}{(z-2)^3}\right]_{z=0}=\left[-\frac{3}{(z-2)^4}\right]_{z=0}=-\frac3{16}.$$
**Residue at $z=2$** (pole of order 3):
$$\text{Res}_{z=2}=\frac1{2!}\frac{d^2}{dz^2}\left[\frac1{z^2}\right]_{z=2}=\frac12\left[\frac6{z^4}\right]_{z=2}=\frac12\cdot\frac6{16}=\frac3{16}.$$
(Check: the residues sum to zero, and $f\sim z^{-5}$ at infinity, so the sum of all residues is 0. ✓)
Answer **$-3/16$ and $3/16$** (option C).