← 2013 Paper 2

UPSC 2013 Maths Optional Paper 2 Q1d — Step-by-Step Solution

10 marks · Section A

Cauchy's residue theorem · Complex Analysis · asked 5× in 13 yrs · Read the full method →

Question

Prove that if $be^{a+1}<1$ where $a$ and $b$ are positive and real, then the function $z^n e^{-a}-be^z$ has $n$ zeros in the unit circle.

Technique

Rouché’s theorem; choose the “dominant” term carefully and bound the “small” perturbation on the boundary.

Solution

Strategy. Apply Rouché’s theorem with $h(z)=z^n e^{-a}$ (which has $n$ zeros at the origin, hence $n$ zeros inside $|z|=1$) and a perturbation $k(z)=-be^z$. Then $h$ and $h+k$ have the same number of zeros inside $|z|=1$ provided $|h(z)|>|k(z)|$ on $|z|=1$.

Step 1 — Bound $|h|$ from below on $|z|=1$

For $|z|=1$:

(Note: $a$ is real, so $e^{-a}$ is just a positive real.)

Step 2 — Bound $|k|$ from above on $|z|=1$

using that $\operatorname{Re}(z)\le|z|$.

Step 3 — Apply the hypothesis

The hypothesis is $be^{a+1}<1$, equivalently

Combined with Steps 1 and 2:

So $|h(z)|>|k(z)|$ on the unit circle.

Step 4 — Rouché’s theorem

Both $h$ and $h+k=z^n e^{-a}-be^z$ are entire (analytic everywhere). On $|z|=1$, $|h|>|k|$. Rouché’s theorem concludes:

since $h(z)=z^n e^{-a}$ has a single zero of multiplicity $n$ at $z=0$.

Answer