← 2014 Paper 2

UPSC 2014 Maths Optional Paper 2 Q1c — Step-by-Step Solution

10 marks · Section A

Cauchy-Riemann equations (necessary and sufficient) · Complex Analysis · asked 5× in 13 yrs · Read the full method →

Question

Prove that the function $f(z)=u+iv$, where

satisfies Cauchy–Riemann equations at the origin, but the derivative of $f$ at $z=0$ does not exist.

Technique

Compute partial derivatives via limit definition at the singular point; verify CR; check path-dependence of the complex derivative.

Solution

Strategy. Compute partials at origin via the limit definition; verify CR; then compute the complex derivative limit along two different paths, finding different limits.

Step 1 — Separate $u$ and $v$

The numerator $x^{3}(1+i)-y^{3}(1-i)=(x^{3}-y^{3})+i(x^{3}+y^{3})$. So

with $u(0,0)=v(0,0)=0$.

Step 2 — Partial derivatives at origin (limit definition)

Step 3 — CR equations at origin

CR equations: $u_x=v_y$ and $u_y=-v_x$.

$u_x(0,0)=1$, $v_y(0,0)=1$. ✓ $u_y(0,0)=-1$, $-v_x(0,0)=-1$. ✓

So CR equations are satisfied at the origin.

Step 4 — Differentiate via path approach

$f'(0)=\lim_{z\to 0}\dfrac{f(z)-f(0)}{z}=\lim_{z\to 0}\dfrac{f(z)}{z}$.

Path 1: $y=0$, $z=x\to 0$. $f(z)=u+iv=\dfrac{x^{3}}{x^{2}}+i\dfrac{x^{3}}{x^{2}}=x+ix=x(1+i)$. $\dfrac{f(z)}{z}=\dfrac{x(1+i)}{x}=1+i$.

Path 2: $y=x$, $z=x+ix=x(1+i)\to 0$. At $y=x$: $u=\dfrac{x^{3}-x^{3}}{2x^{2}}=0$, $v=\dfrac{x^{3}+x^{3}}{2x^{2}}=x$. So $f(z)=ix$. $\dfrac{f(z)}{z}=\dfrac{ix}{x(1+i)}=\dfrac{i}{1+i}=\dfrac{i(1-i)}{2}=\dfrac{1+i}{2}$.

The two limits $1+i$ and $(1+i)/2$ are different — so the path-dependent limit doesn’t have a unique value.

Answer