← 2019 Paper 2

UPSC 2019 Maths Optional Paper 2 Q4b — Step-by-Step Solution

10 marks · Section A

Laurent's series in an annulus · Complex Analysis · asked 7× in 13 yrs · Read the full method →

Question

Obtain the first three terms of the Laurent series expansion of the function $f(z)=\dfrac{1}{(e^z-1)}$ about the point $z=0$ valid in the region $0<|z|<2\pi$.

Technique

Factor $e^z-1=z(1+w)$, geometric expansion of $1/(1+w)$, collect to the needed order, multiply by $1/z$.

Solution

Step 1 — Nature of the singularity

$e^z-1=z+\dfrac{z^2}{2!}+\dfrac{z^3}{3!}+\cdots=z\Big(1+\dfrac{z}{2}+\dfrac{z^2}{6}+\dfrac{z^3}{24}+\cdots\Big)$. It has a simple zero at $z=0$ (and zeros at $z=2\pi i k$), so $f=1/(e^z-1)$ has a simple pole at $0$; the nearest other singularity is at $|z|=2\pi$, giving the annulus $0<|z|<2\pi$.

Step 2 — Factor out $z$ and expand the reciprocal

Let $w=\dfrac{z}{2}+\dfrac{z^2}{6}+\dfrac{z^3}{24}+\cdots$ and use $\dfrac{1}{1+w}=1-w+w^2-w^3+\cdots$ Collect powers of $z$ up to $z^2$:

So

Step 3 — Multiply by $1/z$

Answer