UPSC Maths 2025 Paper 2 Q3a — Solution

15 marks · Section A

Question

Evaluate the integral $\displaystyle\oint_C \dfrac{e^z}{z^2(z+1)^3}\,dz$ , $C : |z| = 2$ .

Technique

Apply the Cauchy residue theorem with higher-order poles: the residue at a pole of order $m$ is $\operatorname{Res}_{z_0} g = \dfrac{1}{(m-1)!}\lim_{z\to z_0}\dfrac{d^{m-1}}{dz^{m-1}}\big[(z-z_0)^m g(z)\big]$ .

Solution

Let $g(z) = \dfrac{e^z}{z^2(z+1)^3}$ . The poles are:

$z=0$ , of order $2$ ,
$z=-1$ , of order $3$ .

Both satisfy $|z|<2$ , so both lie inside $C:|z|=2$ . By the residue theorem, $\oint_C g(z)\,dz = 2\pi i\left[\operatorname{Res}_{z=0} g + \operatorname{Res}_{z=-1} g\right].$

Step 1 — Residue at $z=0$ (order 2). $\operatorname{Res}_{z=0} g = \frac{1}{1!}\lim_{z\to0}\frac{d}{dz}\left[z^2\,g(z)\right] = \lim_{z\to0}\frac{d}{dz}\left[\frac{e^z}{(z+1)^3}\right].$ By the quotient rule, $\frac{d}{dz}\frac{e^z}{(z+1)^3} = \frac{e^z(z+1)^3 - e^z\cdot 3(z+1)^2}{(z+1)^6} = \frac{e^z\big[(z+1)-3\big]}{(z+1)^4} = \frac{e^z(z-2)}{(z+1)^4}.$ At $z=0$ : $\dfrac{e^0(0-2)}{1^4} = -2.$ $\operatorname{Res}_{z=0} g = -2.$

Step 2 — Residue at $z=-1$ (order 3). $\operatorname{Res}_{z=-1} g = \frac{1}{2!}\lim_{z\to-1}\frac{d^2}{dz^2}\left[(z+1)^3\,g(z)\right] = \frac{1}{2}\lim_{z\to-1}\frac{d^2}{dz^2}\left[\frac{e^z}{z^2}\right].$ With $h(z) = e^z z^{-2}$ , $h'(z) = e^z\left(\frac{1}{z^2} - \frac{2}{z^3}\right),$ $h''(z) = e^z\left(\frac{1}{z^2} - \frac{4}{z^3} + \frac{6}{z^4}\right).$ At $z=-1$ : $\ h''(-1) = e^{-1}(1 + 4 + 6) = 11 e^{-1}.$ $\operatorname{Res}_{z=-1} g = \frac{1}{2}\cdot 11 e^{-1} = \frac{11}{2e}.$

Step 3 — Combine. $\oint_C g\,dz = 2\pi i\left(-2 + \frac{11}{2e}\right) = 2\pi i\cdot\frac{11 - 4e}{2e} = \frac{\pi i\,(11 - 4e)}{e} = \pi i\left(\frac{11}{e} - 4\right).$

Numerically $\frac{11}{e}-4 \approx 0.04668$ , so the integral $\approx 0.1466\,i$ .

Answer

$\boxed{\;\oint_{|z|=2} \frac{e^z}{z^2(z+1)^3}\,dz = \pi i\left(\frac{11}{e} - 4\right) = \frac{\pi i\,(11 - 4e)}{e} \approx 0.1466\,i.\;}$