← 2017 Paper 2

UPSC 2017 Maths Optional Paper 2 Q1d — Step-by-Step Solution

10 marks · Section A

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

Question

Determine all entire functions $f(z)$ such that $0$ is a removable singularity of $f\!\left(\dfrac{1}{z}\right)$.

Technique

Riemann removable-singularity theorem (boundedness near $0$) $\to$ boundedness of $f$ at infinity $\to$ boundedness on all of $\mathbb C$ (compactness) $\to$ Liouville $\Rightarrow$ constant. Laurent series gives the same answer directly.

Solution

Write $g(z)=f(1/z)$, which is analytic on the punctured plane $0<|z|<\infty$ (since $f$ is entire and $1/z$ is analytic for $z\ne0$). The condition is that $z=0$ is a removable singularity of $g$. We translate this into a growth condition on $f$ at infinity and apply Liouville’s theorem.

Step 1 — Removable singularity $\Leftrightarrow$ $g$ bounded near $0$

By Riemann’s removable singularity theorem, an isolated singularity $z=0$ of $g$ is removable iff $g$ is bounded on some punctured neighbourhood $0<|z|<\delta$. So the hypothesis is equivalent to:

Step 2 — Translate to behaviour of $f$ at infinity

Substitute $w=1/z$. As $z$ ranges over $0<|z|<\delta$, $w=1/z$ ranges over $|w|>1/\delta=:R$. The bound becomes

Thus $f$ is bounded outside the disc $|w|\le R$.

Step 3 — $f$ is bounded on the whole plane

$f$ is entire, hence continuous on the closed disc $\{|w|\le R\}$, which is compact; so $f$ is bounded there, say $|f(w)|\le M'$ for $|w|\le R$. Combined with Step 2,

So $f$ is a bounded entire function.

Step 4 — Liouville’s theorem

By Liouville’s theorem, a bounded entire function is constant. Hence $f$ is constant.

Conversely, if $f\equiv c$ is constant, then $f(1/z)=c$ is constant on $0<|z|<\infty$, which trivially has a removable singularity at $0$ (the constant extension). So constants exactly satisfy the condition.

Answer