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UPSC 2026 Maths Optional Paper 2 Q1d — Step-by-Step Solution

10 marks · Section A

Cauchy-Riemann equations (necessary and sufficient) · Complex Analysis · asked 6× in 14 yrs · Read the full method →

Question

Show that the function $f(z) = \sqrt{|xy|}$ is not analytic at the origin, although Cauchy-Riemann equations are satisfied at that point.

Technique

This is the standard counterexample to the converse of the Cauchy–Riemann theorem, and the whole mark lies in the contrast, so the answer must do two separate things and keep them separate. First, compute the four partials at the origin from first principles — not by differentiating a formula, because $\sqrt{|xy|}$ has no usable formula-derivative at $(0,0)$ — and observe all four vanish, so CR hold. Second, kill differentiability by the definition: $f'(0)$ is a limit of $\dfrac{f(z)-f(0)}{z}$ over every approach to $0$, and here the substitution $z=re^{i\theta}$ makes the quotient independent of $r$ but dependent on $\theta$ — the limit is different along every ray. The trigger: CR satisfied at a single point is only necessary, never sufficient; sufficiency needs the partials continuous in a neighbourhood.

Solution

Step 0 — Set up the real and imaginary parts.

Write $z=x+iy$ with $x,y\in\mathbb{R}$. The given function is real-valued, so

Note $f(0)=u(0,0)=0$, and $f$ is defined (and continuous) on all of $\mathbb{C}$.

Recall the two things that must not be confused:

• $f$ is differentiable at $z_0$ if $\displaystyle\lim_{z\to z_0}\frac{f(z)-f(z_0)}{z-z_0}$ exists (as a limit in $\mathbb{C}$, i.e. independently of the manner of approach).
• $f$ is analytic (holomorphic) at $z_0$ if it is differentiable at every point of some neighbourhood of $z_0$.

So failure of differentiability at $z_0$ already destroys analyticity at $z_0$.

Step 1 — The four first-order partials at the origin, from the definition.

At the origin the formula $\sqrt{|xy|}$ is not differentiable term-by-term, so the partials must be computed as limits of difference quotients:

(The quotients are identically zero, not merely tending to zero, because $u$ vanishes identically on both coordinate axes.)

Since $v\equiv0$,

Step 2 — The Cauchy–Riemann equations hold at the origin.

So both CR equations

are satisfied at $z=0$, and all four partials exist there. $\qquad\checkmark$

Step 3 — Yet $f'(0)$ does not exist: the difference quotient depends on the direction.

By definition,

Put $z=re^{i\theta}$ with $r>0$, so $x=r\cos\theta$, $y=r\sin\theta$ and $|xy|=r^{2}\left|\cos\theta\sin\theta\right|$. Since $r>0$, $\sqrt{|xy|}=r\sqrt{\left|\cos\theta\sin\theta\right|}$, and hence

This is a remarkable and decisive fact: the quotient does not depend on $r$ at all. Along the ray of direction $\theta$ it is the constant $\Phi(\theta)$, so the limit along that ray exists and equals $\Phi(\theta)$. Two rays now settle the matter:

• along $\theta=0$ (the positive real axis): $\ \Phi(0)=\sqrt{|1\cdot0|}\,e^{0}=0$;
• along $\theta=\dfrac{\pi}{4}$ (the line $y=x$, $x>0$): $\ \Phi\!\left(\tfrac{\pi}{4}\right)=\sqrt{\left|\tfrac{1}{\sqrt2}\cdot\tfrac{1}{\sqrt2}\right|}\;e^{-i\pi/4}=\frac{1}{\sqrt2}\cdot\frac{1-i}{\sqrt2}=\frac{1-i}{2}.$

Since

the limit $\displaystyle\lim_{z\to0}\frac{f(z)-f(0)}{z}$ does not exist. Hence $f$ is not differentiable at $z=0$, and a fortiori not analytic at $z=0$. $\qquad\blacksquare$

(Equivalently in Cartesian form: approaching along $y=mx$, $x\to0^{+}$, gives $\dfrac{\sqrt{|m|}\,x}{x(1+im)}=\dfrac{\sqrt{|m|}}{1+im}$, which is $0$ for $m=0$ and $\dfrac{1}{1+i}=\dfrac{1-i}{2}$ for $m=1$ — the same conclusion. The polar form is preferable because it exhibits the $r$-independence, showing the failure is purely directional.)

Step 4 — Why this is no contradiction: exactly which hypothesis fails.

The theorem that CR “implies” differentiability is a sufficient condition with an extra hypothesis:

If $u,v$ have first-order partials in a neighbourhood of $z_{0}$, those partials are continuous at $z_{0}$, and CR hold at $z_{0}$, then $f'(z_{0})$ exists.

(Equivalently and more sharply: if $u,v$ are real-differentiable at $z_{0}$ and CR hold at $z_{0}$, then $f'(z_{0})$ exists.)

Here the extra hypothesis fails, and it is worth showing precisely how.

(a) $u$ is not real-differentiable at $(0,0)$. If it were, then since $u_{x}(0,0)=u_{y}(0,0)=0$ the differential would be the zero map, forcing

But along $y=x>0$,

So $u$ fails to be differentiable at the origin as a function of two real variables — this is the precise defect.

(b) The partials are not continuous at $(0,0)$ — indeed they are unbounded near it. For $x>0,\ y>0$ we have $u=\sqrt{xy}$, so

while $u_{x}(0,0)=0$.

So CR holding at a point carries no differentiability information whatever; the CR equations are necessary but not sufficient, and this function is the textbook witness.

Step 5 — A stronger conclusion: $f$ is analytic nowhere.

Analyticity at $0$ would require differentiability, hence CR, at every point of a whole neighbourhood of $0$. In fact the CR equations hold only at the origin:

• If $a\neq0$ and $b\neq0$, then near $(a,b)$ the sign of $xy$ is constant and $u=\sqrt{|x||y|}$ is smooth, with

whereas CR would demand $u_{x}=v_{y}=0$. CR fail.

• If $b=0$, $a\neq0$: $\ \dfrac{u(a,k)-u(a,0)}{k}=\dfrac{\sqrt{|a|}\sqrt{|k|}}{k}=\pm\dfrac{\sqrt{|a|}}{\sqrt{|k|}}\to\pm\infty$, so $u_{y}(a,0)$ does not exist; CR cannot even be stated, let alone hold.
• Symmetrically, if $a=0$, $b\neq0$, then $u_{x}(0,b)$ does not exist.

Hence every punctured neighbourhood of $0$ contains points where CR fail, confirming again that $f$ is not analytic at $0$ — and that it is not analytic at any point of $\mathbb{C}$.

(Consistency remark, one line: a real-valued function analytic on a domain is necessarily constant there, since $v\equiv0$ forces $u_{x}=v_{y}=0$ and $u_{y}=-v_{x}=0$. So no non-constant real-valued $f$, this one included, can be analytic anywhere — the most a function like $\sqrt{|xy|}$ can achieve is differentiability at isolated points, and even that fails here.)

Answer