UPSC 2024 Maths Optional Paper 2 Q2c — Step-by-Step Solution

20 marks · Section A

Analytic Functions: Complex Differentiability · Complex Analysis · Read the full method →

Question

Find the function which is analytic inside and on the circle $C:z=e^{i\theta}$ , $0\le\theta\le 2\pi$ , and takes the value

\frac{(a^2-1)\cos\theta+i(a^2+1)\sin\theta}{a^4-2a^2\cos 2\theta+1}

on the circumference, where $a^2>1$ .

Technique

Substitute $z=e^{i\theta}$ to write $\cos\theta$ , $\sin\theta$ , $\cos 2\theta$ as rational functions of $z$ ; algebraic simplification reveals the analytic function.

Solution

Step 1 — Rewrite the boundary value using $z=e^{i\theta}$ .

On $|z|=1$ : $\cos\theta=\tfrac{z+z^{-1}}{2}$ , $\sin\theta=\tfrac{z-z^{-1}}{2i}$ , $\cos 2\theta=\tfrac{z^2+z^{-2}}{2}$ .

Numerator:

(a^2-1)\cos\theta+i(a^2+1)\sin\theta=\frac{(a^2-1)(z+z^{-1})}{2}+\frac{(a^2+1)(z-z^{-1})}{2} =\frac{2a^2 z-2z^{-1}}{2}=a^2 z-z^{-1}=\frac{a^2 z^2-1}{z}.

Denominator: $a^4-2a^2\cos 2\theta+1=a^4-a^2(z^2+z^{-2})+1$ . Multiply by $z^2$ :

z^2\cdot\text{denom}=a^4 z^2-a^2 z^4-a^2+z^2=-(a^2 z^2-1)(z^2-a^2).

So denom $=-(a^2 z^2-1)(z^2-a^2)/z^2$ .

Step 2 — Combine.

\text{Boundary value}=\frac{(a^2 z^2-1)/z}{-(a^2 z^2-1)(z^2-a^2)/z^2}=\frac{-z}{z^2-a^2}=\frac{z}{a^2-z^2}.

Step 3 — Verify analyticity on $|z|\le 1$ .

The poles of $f(z)=z/(a^2-z^2)$ are at $z=\pm a$ . Since $a^2>1$ , $|a|>1$ : both poles lie outside the closed unit disk. So $f$ is analytic on $|z|\le 1$ .

By uniqueness of analytic continuation, this is the required function.

Answer

\boxed{\;f(z)=\frac{z}{a^2-z^2}.\;}