Let f : ℂ → ℂ be a real-differentiable function and define u(x, y) = Re f(x + iy), v(x, y) = Im f(x + iy). Let ∇u = (uₓ, uy) denote the gradient. Which one of the following is necessarily true?

C. For entire f the Cauchy–Riemann equations give ∇u · ∇v = uₓvₓ + uyvy = uₓ(−uy) + uyuₓ = 0. Without holomorphy there is no such relation, and orthogonal gradients alone do not force CR (e.g. f = z̄ has ∇u · ∇v = 0).

Cr orthogonality — CSIR NET December 2023 Part B, solved

Cr orthogonality

Let $f : \mathbb{C} \to \mathbb{C}$ be a real-differentiable function and define u(x, y) = Re f(x + iy), v(x, y) = Im f(x + iy). Let $\nabla{}u = (u_{x}, u_y)$ denote the gradient. Which one of the following is necessarily true?

A.For $c_{1}, c_{2} \in \mathbb{C}$ , the level curves $u = c_{1}$ and $v = c_{2}$ are orthogonal wherever they intersect.
B. $\nabla{}u \cdot \nabla{}v = 0$ at every point.
C.If f is an entire function, then $\nabla{}u \cdot \nabla{}v = 0$ at every point.✓ Correct
D.If $\nabla{}u \cdot \nabla{}v = 0$ at every point, then f is an entire function.

The idea, in plain words

When a complex function is genuinely complex-differentiable, its real and imaginary parts are locked together by a pair of equations, and those equations force the two gradients to point at right angles. Merely being differentiable as a function of two real variables buys you none of that locking.

Solution

For entire f the Cauchy–Riemann equations give $\nabla{}u \cdot \nabla{}v = u_{x}v_{x} + u_yv_y = u_{x}(-u_y) + u_yu_{x} = 0$ .

Without holomorphy there is no such relation, and orthogonal gradients alone do not force CR (e.g. f = z̄ has $\nabla{}u \cdot \nabla{}v = 0)$ .

Why each option is right or wrong

Checked against 2 symbolic computations

A.WrongHypothesis dropped
False. f is only assumed real-differentiable

so the Cauchy–Riemann equations need not hold and there is no reason for the level curves to meet at right angles.

Even for entire f the statement is delicate at points where f′ = 0

since the level sets there need not be curves at all.

The hypothesis granted is much weaker than the one the claim needs.
B.WrongHypothesis dropped
False for exactly the same reason: without holomorphy, u and v are unrelated real functions and their gradients can point anywhere.

Take f(x + iy) = x + ix

so u = v = x and $\nabla{}u \cdot \nabla{}v = 1 \ne 0$ .

Option (C) is this same claim with the missing hypothesis restored.
C.Correct
True.

For entire f the Cauchy–Riemann equations give $u_{x} = v_y$ and $u_y = -v_{x}$ .

Hence $\nabla{}u \cdot \nabla{}v = u_{x}v_{x} + u_yv_y = u_{x}(-u_y) + u_y(u_{x}) = 0$ identically.

SymPy checks the model case $f(z) = z^{2}$ , where $u = x^{2} - y^{2}$ and v = 2xy: the dot product simplifies to 0.

Note this holds everywhere

including at zeros of f′ — the gradients may vanish

but the product is still 0.
D.WrongConverse assumed
False — it is (C) read backwards.

Take f(z) = z̄

so u = x and v = −y: SymPy gives $\nabla{}u \cdot \nabla{}v = 1\cdot0 + 0\cdot(-1) = 0$

yet f is nowhere holomorphic.

Orthogonal gradients are one consequence of Cauchy–Riemann, not equivalent to them.

3 of these 3 wrong options correspond to a named reasoning error — the same errors recur across subjects.

The trap it tests

Execution slip

The idea was right. The computation was not.

Drill statements like this

Last revised 12 Sept 2026. Found a mistake? Tell us — corrections are the fastest thing we act on.

Cr orthogonality

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