← 2021 Paper 2

UPSC 2021 Maths Optional Paper 2 Q3a — Step-by-Step Solution

15 marks · Section A

Singularities: removable, pole, essential · Complex Analysis · asked 3× in 13 yrs · Read the full method →

Question

Let $f$ be entire with Taylor series at $z=0$ having infinitely many terms. Show $z=0$ is an essential singularity of $f(1/z)$.

Technique

Direct: substitute $z\to 1/z$ in Taylor series; the resulting Laurent series at $0$ has infinitely many negative-power terms; classify as essential.

Solution

Setup. $f$ entire ⇒ Taylor series at $0$ converges everywhere: $f(z)=\sum_{n=0}^\infty a_n z^n$, with infinitely many $a_n\ne 0$.

Step 1 — Substitute $z\to 1/z$

$g(z):=f(1/z)=\sum_{n=0}^\infty a_n z^{-n}=a_0+\dfrac{a_1}{z}+\dfrac{a_2}{z^2}+\cdots$.

This is the Laurent series of $g$ centred at $z=0$, valid for $|z|>0$ (since $f(w)$ is defined for all $w=1/z\ne 0$, i.e., $z\ne 0$).

Step 2 — Classify singularity at $z=0$

A singularity at $z_0$ has type determined by the principal part of the Laurent series (terms with negative powers):

• Removable: principal part has no terms.
• Pole: principal part has finitely many (nonzero) terms.
• Essential: principal part has infinitely many nonzero terms.

For $g(z)=\sum a_n z^{-n}$, the negative-power terms are $a_n/z^n$ for $n\ge 1$. Since infinitely many $a_n\ne 0$, infinitely many of these terms are nonzero.

Therefore $z=0$ is an essential singularity of $g(z)=f(1/z)$. $\blacksquare$

Answer