← 2022 Paper 2

UPSC 2022 Maths Optional Paper 2 Q1b — Step-by-Step Solution

10 marks · Section A

Harmonic functions and harmonic conjugate · Complex Analysis · asked 7× in 13 yrs · Read the full method →

Question

If $f(z)=u+iv$ is analytic and $u-v=\dfrac{\cos x+\sin x-e^{-y}}{2\cos x-e^y-e^{-y}}$, find $f(z)$ subject to $f(\pi/2)=0$.

Technique

Milne-Thomson method on $g(z)=(1-i)f(z)$ where $\Re g=u-v$; restrict to $y=0$, simplify with half-angle identities; analytically continue to recover $g(z)$.

Solution

Strategy. Use Milne-Thomson’s method: if $f(z)=u+iv$ is analytic, then $f'(z)=u_x+iv_x=u_x-iu_y$ (using CR equations).

Let $\psi(x,y)=u-v$. We can construct $(1-i)f(z)=u-v+i(u+v)$. So define $g(z)=(1-i)f(z)$, with $\Re g=u-v$ and $\Im g=u+v$.

Alternative cleaner approach: Let $U=u-v$, $V=u+v$ (the real and imaginary parts of $(1-i)f$). The function $(1-i)f$ is analytic, so $U$ is harmonic and $V$ is its conjugate.

Given $U=\dfrac{\cos x+\sin x-e^{-y}}{2\cos x-e^y-e^{-y}}$.

Simplify denominator: $2\cos x-e^y-e^{-y}=2\cos x-2\cosh y=2(\cos x-\cosh y)$.

Numerator: $\cos x+\sin x-e^{-y}$.

So $U=\dfrac{\cos x+\sin x-e^{-y}}{2(\cos x-\cosh y)}$.

Step 1 — Apply Milne-Thomson to $g(z)=(1-i)f(z)$

$g'(z)=U_x-i U_y$ (using CR on $g$).

We need $U_x,U_y$ at $y=0$ (where the formula reduces to a function of $x$ alone).

At $y=0$: $\cosh 0=1$, $e^{0}=1$. So $U(x,0)=\dfrac{\cos x+\sin x-1}{2(\cos x-1)}$.

Simplify: numerator $=\cos x-1+\sin x=-2\sin^2(x/2)+2\sin(x/2)\cos(x/2)=2\sin(x/2)[\cos(x/2)-\sin(x/2)]$.

Denominator: $2(\cos x-1)=2[-2\sin^2(x/2)]=-4\sin^2(x/2)$.

$U(x,0)=\dfrac{2\sin(x/2)[\cos(x/2)-\sin(x/2)]}{-4\sin^2(x/2)}=\dfrac{\cos(x/2)-\sin(x/2)}{-2\sin(x/2)}=\dfrac{\sin(x/2)-\cos(x/2)}{2\sin(x/2)}=\dfrac{1}{2}-\dfrac{\cos(x/2)}{2\sin(x/2)}=\dfrac{1}{2}-\dfrac{1}{2}\cot(x/2)$.

So $U(x,0)=\dfrac{1}{2}-\dfrac{1}{2}\cot(x/2)$.

Step 2 — Construct $g(z)$ via Milne-Thomson

Since $g(z)$ on the real axis ($y=0$) equals $U(x,0)+iV(x,0)$, and we want to identify the analytic function whose real part on the real axis matches $U(x,0)$.

By Milne-Thomson: $g(z)=U(z,0)-i\int U_y(z,0)\,dz$… but actually the simpler version is:

If $U(x,y)=\Re g$, replace $x\to z,\;y\to 0$ to get $g(z)$ on the real axis. For analytic functions, the formula extends.

$g(z)=U(z,0)$ formally (where the formula $U(x,0)$ is analytically continued to complex $z$).

$g(z)=\dfrac{1}{2}-\dfrac{1}{2}\cot(z/2)$.

(Note: this gives $g$ up to an additive imaginary constant, which we’ll fix using the initial condition.)

Step 3 — Recover $f(z)$

$g(z)=(1-i)f(z)\Rightarrow f(z)=\dfrac{g(z)}{1-i}=\dfrac{g(z)(1+i)}{2}$.

Step 4 — Apply initial condition $f(\pi/2)=0$

$f(\pi/2)=\dfrac{1+i}{4}[1-\cot(\pi/4)]=\dfrac{1+i}{4}[1-1]=0$ ✓.

So the additive constant is $0$:

Answer