← 2023 Paper 2

UPSC 2023 Maths Optional Paper 2 Q4b — Step-by-Step Solution

15 marks · Section A

Singularities: removable, pole, essential · Complex Analysis · asked 3× in 14 yrs · Read the full method →

Question

Classify the singular point $z=0$ of the function $f(z)=\dfrac{e^z}{z-\sin z}$ and obtain the principal part of its Laurent series expansion.

Technique

Power-series expansion of the denominator; geometric-series inversion; multiply by $e^{z}$ and gather negative-power terms.

Solution

Step 1 — Order of the zero of $z-\sin z$ at $0$

Using the Maclaurin series $\sin z=z-\dfrac{z^{3}}{3!}+\dfrac{z^{5}}{5!}-\cdots$:

The leading term is $z^{3}/6$, so $z-\sin z$ has a zero of order 3 at $z=0$.

Step 2 — Classify the singularity

$e^{z}$ is entire with $e^{0}=1\ne 0$. The ratio $f(z)=\dfrac{e^{z}}{z-\sin z}$ therefore has a pole of order 3 at $z=0$.

Step 3 — Series of $\dfrac{1}{z-\sin z}$ near $z=0$

Write

Using $\dfrac{1}{1-u}=1+u+u^{2}+O(u^{3})$ and retaining terms up to $z^{4}$:

(Detail: $-\dfrac{1}{840}+\dfrac{1}{400}=\dfrac{-400+840}{840\cdot 400}=\dfrac{440}{336000}=\dfrac{11}{8400}$.)

Therefore

Step 4 — Multiply by $e^{z}$ and collect principal part

$e^{z}=1+z+\dfrac{z^{2}}{2}+\dfrac{z^{3}}{6}+\dfrac{z^{4}}{24}+\cdots$.

Multiplying $(\ast)$ by $e^{z}$ and collecting terms with negative powers of $z$:

• Coefficient of $z^{-3}$: $\dfrac{6}{z^{3}}\cdot 1=\dfrac{6}{z^{3}}$.
• Coefficient of $z^{-2}$: $\dfrac{6}{z^{3}}\cdot z=\dfrac{6}{z^{2}}$.
• Coefficient of $z^{-1}$: $\dfrac{6}{z^{3}}\cdot\dfrac{z^{2}}{2}+\dfrac{3}{10\,z}\cdot 1=\dfrac{3}{z}+\dfrac{3}{10\,z}=\dfrac{33}{10\,z}$.

The principal part contains exactly these three terms:

Answer