Cauchy-Riemann equations (necessary and sufficient)

At a Glance

• Frequency: 5 sub-parts across 5 of 13 years (2014, 2019, 2020, 2023, 2024)
• Priority tier: T2
• Marks (count): 10 (4), 15 (1)
• Average solve time: ~8 min
• Difficulty mix: easy 3, medium 2
• Section: A | Dominant type: proof

Why This Chapter Matters

The Cauchy-Riemann equations are the single most-tested idea in complex analysis at this level: every year from 2019 to 2024 carried a CR question. They appear in four flavours — showing CR hold but the derivative fails, proving an analytic function must be constant, recovering an analytic function from one harmonic part, and verifying analyticity using CR directly — and a candidate who can execute all four is essentially guaranteed full marks on any Section A complex-analysis part. The polar form of the CR equations, often overlooked, is indispensable for the recovery archetype and for $\log z$.

Minimum Theory

CR equations (Cartesian). If $f = u + iv$ is differentiable at $z_0 = x_0 + iy_0$ then the partial derivatives satisfy $u_x = v_y, \qquad u_y = -v_x \quad \text{at } (x_0, y_0).$ These are necessary for differentiability. Alone they are not sufficient: the classic pathology (2014) shows CR can hold at a point while $f'$ fails to exist.

Sufficient conditions (Milne-Thomson theorem). If $u$ and $v$ have continuous first-order partial derivatives on an open neighbourhood of $z_0$ and the CR equations hold at $z_0$, then $f$ is analytic at $z_0$ and $f'(z) = u_x + iv_x = v_y - iu_y.$ The continuity of partials in a neighbourhood (not just at the point) is the extra ingredient that converts CR into analyticity.

CR equations (Polar). With $z = re^{i\theta}$, $u = u(r,\theta)$, $v = v(r,\theta)$: $u_r = \frac{1}{r}\,v_\theta, \qquad u_\theta = -r\,v_r, \qquad f'(z) = e^{-i\theta}(u_r + iv_r).$ Use these when the given harmonic part is expressed in $(r,\theta)$ — the recovery archetype always arrives in polar form.

Harmonic conjugates. If $f = u + iv$ is analytic on a domain, then $\nabla^2 u = 0$ and $\nabla^2 v = 0$; $v$ is called a harmonic conjugate of $u$. Conversely, given a harmonic function $u$ (or $v$), a conjugate can be recovered by integrating the CR relations.

Question Archetypes

cr-pathology (1 question; 2014)

Show the Cauchy-Riemann equations hold at a point yet the derivative fails to exist.

Recognition Cues

The question gives a piecewise-defined $f$ (often $f(0)=0$ with a rational formula elsewhere) and asks to verify CR at the origin and show that $f'(0)$ does not exist. The key signal is that two limits of $f(z)/z$ along different paths disagree.

Solution Template

1. Write $f = u + iv$ and read off $u$ and $v$ from the formula.
2. Compute $u_x(0,0)$ and $u_y(0,0)$ using the limit definition: $u_x(0,0) = \lim_{h\to0}\frac{u(h,0)-u(0,0)}{h}, \quad u_y(0,0) = \lim_{k\to0}\frac{u(0,k)-u(0,0)}{k}.$ Do the same for $v_x$ and $v_y$. Confirm $u_x = v_y$ and $u_y = -v_x$.
3. Test $f'(0) = \lim_{z\to0} f(z)/z$ along two different paths (e.g., real axis $y=0$ and diagonal $y=x$). Obtain two different limits.
4. Conclude: CR necessary conditions hold, but the limit is path-dependent, so $f'(0)$ does not exist. State that the partials are not continuous in any neighbourhood of $0$.

Worked Example

2014 Paper 2, 2014-P2-Q1c (10 marks)

Let $f(z) = \dfrac{x^3(1+i)-y^3(1-i)}{x^2+y^2}$ for $z\ne0$ and $f(0)=0$. Show CR equations are satisfied at $z=0$ but $f'(0)$ does not exist.

Separate real and imaginary parts: $u(x,y) = \frac{x^3 - y^3}{x^2+y^2}, \qquad v(x,y) = \frac{x^3+y^3}{x^2+y^2}.$

CR at the origin (limit definition): $u_x(0,0) = \lim_{h\to0}\frac{h^3/h^2}{h} = \lim_{h\to0} 1 = 1.$ $u_y(0,0) = \lim_{k\to0}\frac{-k^3/k^2}{k} = \lim_{k\to0}(-1) = -1.$ $v_x(0,0) = \lim_{h\to0}\frac{h^3/h^2}{h} = 1, \qquad v_y(0,0) = \lim_{k\to0}\frac{k^3/k^2}{k} = 1.$

Check: $u_x = v_y = 1$ \checkmark, $\quad u_y = -v_x = -1$ \checkmark. CR hold at the origin.

Derivative fails to exist:

Along the real axis ($y=0$): $\frac{f(z)}{z} = \frac{x\cdot\frac{x^3(1+i)}{x^2}}{x} = 1+i.$

Along the diagonal ($y=x$, $z = x(1+i)$): $\frac{f(z)}{z} = \frac{\frac{x^3(1+i)-x^3(1-i)}{2x^2}}{x(1+i)} = \frac{x\cdot 2i/2}{x(1+i)} = \frac{i}{1+i} = \frac{1+i}{2}.$

Since $1+i \ne \dfrac{1+i}{2}$, the limit of $f(z)/z$ as $z\to0$ is path-dependent. Therefore $f'(0)$ does not exist.

$\boxed{\text{CR hold at }z=0,\text{ but }f'(0)\text{ does not exist.}}$

Common Traps

• Do not compute partials by the quotient rule at $(0,0)$. The formula $f(z)/(x^2+y^2)$ is not differentiable at the origin by inspection; you must use the limit definition, otherwise you are dividing by zero.
• One path is not enough. Showing $f(z)/z\to 1+i$ along the real axis only says the limit exists along that path; you need a second path that gives a different value to prove non-existence.
• “CR at a single point” is the entire point of this archetype. The partials are discontinuous in every punctured neighbourhood of $0$, which is why the Milne-Thomson sufficient condition fails.

analytic-constant (1 question; 2019)

Show that an analytic function satisfying a relation between $\operatorname{Re}f$ and $\operatorname{Im}f$ must be constant.

Recognition Cues

The question states ”$f$ is analytic on a domain $D$” and imposes a constraint such as $v = u^2$, $v = g(u)$, $|f| = \text{const}$, or $\arg f = \text{const}$. The instruction is “show $f$ is constant.” The method is always: differentiate the constraint, substitute CR, factor into a product $(\text{positive factor}) \cdot (\partial) = 0$, conclude the partials are zero, hence $f$ is constant on the connected domain.

Solution Template

1. Write the constraint explicitly: e.g. $v = u^2$.
2. Differentiate in $x$ and $y$ to get $v_x, v_y$ in terms of $u, u_x, u_y$.
3. Substitute CR ($u_x = v_y$, $u_y = -v_x$) to obtain a system purely in $u, u_x, u_y$.
4. Eliminate one variable: substitute the first equation into the second to obtain $P \cdot u_x = 0$ where $P > 0$ everywhere.
5. Conclude $u_x = 0$, back-substitute to get $u_y = v_x = v_y = 0$.
6. State: all partials vanish on connected $D$ $\Rightarrow$ $u, v$ constant $\Rightarrow$ $f$ constant.

Worked Example

2019 Paper 2, 2019-P2-Q1d (10 marks)

Let $f = u + iv$ be analytic on a domain $D$ and $v = u^2$ throughout $D$. Prove $f$ is constant.

Differentiate $v = u^2$: $v_x = 2u \cdot u_x, \qquad v_y = 2u \cdot u_y.$

Apply CR ($u_x = v_y$, $u_y = -v_x$): $u_x = v_y = 2u \cdot u_y, \qquad u_y = -v_x = -2u \cdot u_x.$

Substitute the first equation into the second: $u_x = 2u \cdot u_y = 2u\cdot(-2u \cdot u_x) = -4u^2 u_x,$ $u_x(1 + 4u^2) = 0.$

Since $1 + 4u^2 \geq 1 > 0$ for all real $u$, we must have $u_x = 0$.

Then $u_y = -2u \cdot u_x = 0$, and from $v_x = 2u u_x = 0$, $v_y = 2u u_y = 0$.

All four partial derivatives vanish on the connected domain $D$, so $u$ and $v$ are individually constant, hence

$\boxed{f \text{ is constant on } D.}$

Common Traps

• Do not divide by $u$. If $u = 0$ somewhere the division is invalid; instead factor out $u_x$ to get $(1+4u^2)u_x = 0$ and use the positivity of $1+4u^2$.
• Connectedness is essential. All-zero partials imply constancy only on a connected domain. If $D$ is disconnected, $f$ is only piecewise constant. State “connected domain” in the conclusion.
• Check both CR equations are used. The proof uses $u_x = v_y$ and $u_y = -v_x$; skipping one leaves the system underdetermined and the examiner will deduct marks.

recover-analytic-polar (1 question; 2020)

Given one harmonic part in polar coordinates, recover the full analytic function $f(z)$.

Recognition Cues

The question gives $v(r,\theta)$ (or $u(r,\theta)$) in polar form and asks to find $f(z) = u + iv$. The signal words are “find an analytic function” or “determine $f(z)$.” Since the given part contains $r^{\pm1}$ and $\sin\theta/\cos\theta$, the polar CR equations are the correct tool; using Cartesian CR leads to messy algebra.

Solution Template

1. Compute $v_r$ and $v_\theta$ from the given $v(r,\theta)$.
2. Use polar CR: $u_r = \tfrac{1}{r}v_\theta$ and $u_\theta = -r v_r$.
3. Integrate $u_r$ with respect to $r$, holding $\theta$ fixed: $u = \int u_r\,dr + g(\theta)$.
4. Differentiate w.r.t. $\theta$ and match with $u_\theta$ from step 2 to find $g'(\theta)$, then $g(\theta)$ (usually a constant $C$).
5. Write $f = u + iv$ and recognise the combination as a known function of $z$ (often $z + 1/z$, $z^n$, etc.), absorbing $C$.

Worked Example

2020 Paper 2, 2020-P2-Q4a (15 marks)

If $v(r,\theta) = \left(r - \dfrac{1}{r}\right)\sin\theta$, find the analytic function $f(z) = u + iv$.

Compute $v_r$ and $v_\theta$: $v_r = \left(1 + \frac{1}{r^2}\right)\sin\theta, \qquad v_\theta = \left(r - \frac{1}{r}\right)\cos\theta.$

Apply polar CR to find $u_r$: $u_r = \frac{1}{r}\,v_\theta = \frac{1}{r}\left(r-\frac{1}{r}\right)\cos\theta = \left(1 - \frac{1}{r^2}\right)\cos\theta.$

Integrate $u_r$ with respect to $r$: $u = \int\left(1 - \frac{1}{r^2}\right)\cos\theta\,dr = \left(r + \frac{1}{r}\right)\cos\theta + g(\theta).$

Find $g(\theta)$ via $u_\theta$: $u_\theta = -\left(r+\frac{1}{r}\right)\sin\theta + g'(\theta).$ Polar CR also gives $u_\theta = -r v_r = -r\left(1+\dfrac{1}{r^2}\right)\sin\theta = -\left(r+\dfrac{1}{r}\right)\sin\theta$.

Matching: $g'(\theta) = 0$, so $g(\theta) = C$ (constant).

Assemble $f$: $f = \left(r+\frac{1}{r}\right)\cos\theta + i\left(r-\frac{1}{r}\right)\sin\theta + C.$

Recognise using $e^{\pm i\theta} = \cos\theta \pm i\sin\theta$: $f = r e^{i\theta} + \frac{1}{r}e^{-i\theta} + C = z + \frac{1}{z} + C.$

$\boxed{f(z) = z + \frac{1}{z} + C.}$

Common Traps

• Use polar CR, not Cartesian. Converting $v(r,\theta)$ to Cartesian and then applying $u_x = v_y$ is algebraically correct but extremely messy; polar CR is the designed tool.
• Do not forget $g(\theta)$. Integrating $u_r$ in $r$ leaves an arbitrary function of $\theta$ that must be determined. Omitting this is the most common error.
• Recognise the closed form. After assembling $u + iv$, factor out $re^{i\theta}$ and $(1/r)e^{-i\theta}$ to write the answer as an elementary function of $z$; leaving it in $(r,\theta)$ form loses the elegance and may cost presentation marks.

analyticity-conditions (1 question; 2023)

State the sufficient conditions for analyticity, verify them for $\log z$, and compute the derivative.

Recognition Cues

The question explicitly asks to “state sufficient conditions” before the application, signalling that a theorem statement is required. The application is typically $\log z$, $z^n$, or a rational function, and always involves identifying a branch cut (the standard principal branch of $\log z$ has a cut along $(-\infty, 0]$).

Solution Template

1. State the theorem: If $u$, $v$ have continuous first-order partial derivatives in an open neighbourhood of $z_0$, and if the CR equations $u_x = v_y$, $u_y = -v_x$ hold at $z_0$, then $f = u + iv$ is analytic at $z_0$ and $f'(z_0) = u_x + iv_x$.
2. Write the given function in $u + iv$ form; specify the domain (eliminate the branch cut).
3. Compute all four partials $u_x, u_y, v_x, v_y$.
4. Verify CR hold identically on the domain.
5. Check continuity (usually obvious for rational expressions in $x, y$).
6. Apply the formula $f' = u_x + iv_x$ and simplify to a function of $z$.

Worked Example

2023 Paper 2, 2023-P2-Q1d (10 marks)

State sufficient conditions for a function $f(z) = u + iv$ to be analytic. Show that $f(z) = \log z$ is analytic for $z \ne 0$ with a suitable branch cut, and find $df/dz$.

Theorem statement. If $u_x, u_y, v_x, v_y$ exist and are continuous in an open neighbourhood of $z_0$, and the CR equations $u_x = v_y$, $u_y = -v_x$ hold at $z_0$, then $f$ is analytic at $z_0$ with $f'(z_0) = u_x + iv_x$.

Apply to $\log z$. Take the principal branch: $\log z = \ln|z| + i\arg z$ on $\mathbb{C} \setminus (-\infty, 0]$.

$u = \ln\sqrt{x^2+y^2} = \tfrac{1}{2}\ln(x^2+y^2), \qquad v = \arctan\!\left(\frac{y}{x}\right) \text{ (with appropriate quadrant adjustment)}.$

Partial derivatives: $u_x = \frac{x}{x^2+y^2}, \quad u_y = \frac{y}{x^2+y^2}, \quad v_x = \frac{-y}{x^2+y^2}, \quad v_y = \frac{x}{x^2+y^2}.$

Verify CR: $u_x = v_y = \frac{x}{x^2+y^2} \checkmark, \qquad u_y = -v_x = \frac{y}{x^2+y^2} \checkmark.$

All four partials are continuous on $\mathbb{C}\setminus\{0\}$ away from the branch cut. By the theorem, $\log z$ is analytic on $\mathbb{C}\setminus(-\infty,0]$.

Derivative: $\frac{d}{dz}\log z = u_x + iv_x = \frac{x}{x^2+y^2} + i\cdot\frac{-y}{x^2+y^2} = \frac{x-iy}{x^2+y^2} = \frac{\bar{z}}{|z|^2} = \frac{1}{z}.$

$\boxed{\frac{d}{dz}\log z = \frac{1}{z}.}$

Common Traps

• CR alone is not sufficient. A common exam error is stating only “CR hold $\Rightarrow$ analytic.” The theorem requires both CR and continuity of the partials in a neighbourhood. State both conditions.
• Specify the branch cut. $\log z$ is multivalued; without specifying the principal branch cut along $(-\infty,0]$, the argument that $v$ is a well-defined real function is incomplete. Examiners expect the domain to be stated.
• Use the $f' = u_x + iv_x$ formula directly. Computing $d(\log z)/dz$ from scratch (e.g., using $d/dz = e^{-i\theta}(\partial_r)$) is longer than just evaluating $u_x + iv_x$ and simplifying to $1/z$.

verify-cr (1 question; 2024)

Verify the CR equations for $f = p + iq$ where $p$ and $q$ are defined in terms of two harmonic functions.

Recognition Cues

The question defines $p$ and $q$ explicitly as linear combinations of partial derivatives of given harmonic functions $\varphi$ and $\psi$, and asks to show $f = p + iq$ is analytic. The harmonic condition ($\nabla^2\varphi = 0$, $\nabla^2\psi = 0$) is what makes the two CR equations close.

Solution Template

1. Write out $p_x$, $p_y$, $q_x$, $q_y$ by differentiating the given expressions for $p$ and $q$.
2. Form $p_x - q_y$ and $p_y + q_x$; group terms to isolate $\nabla^2\varphi$ and $\nabla^2\psi$.
3. Use $\varphi_{xx}+\varphi_{yy}=0$ and $\psi_{xx}+\psi_{yy}=0$ to conclude both expressions equal zero.
4. State: $p_x = q_y$ and $p_y = -q_x$ (CR), partials continuous (since $\varphi,\psi$ are harmonic hence smooth), therefore $f = p + iq$ is analytic.

Worked Example

2024 Paper 2, 2024-P2-Q1d (10 marks)

If $\varphi$ and $\psi$ are harmonic functions of $(x,y)$, show that $f = p + iq$ is analytic where $p = \varphi_y - \psi_x$ and $q = \varphi_x + \psi_y$.

Compute partial derivatives of $p$ and $q$: $p_x = \varphi_{yx} - \psi_{xx}, \quad p_y = \varphi_{yy} - \psi_{xy},$ $q_x = \varphi_{xx} + \psi_{yx}, \quad q_y = \varphi_{xy} + \psi_{yy}.$

First CR equation ($p_x = q_y$): $p_x - q_y = \varphi_{yx} - \psi_{xx} - \varphi_{xy} - \psi_{yy} = -(\psi_{xx}+\psi_{yy}) = 0$ since $\psi$ is harmonic. Hence $p_x = q_y$ \checkmark.

Second CR equation ($p_y = -q_x$): $p_y + q_x = \varphi_{yy} - \psi_{xy} + \varphi_{xx} + \psi_{xy} = \varphi_{xx}+\varphi_{yy} = 0$ since $\varphi$ is harmonic. Hence $p_y = -q_x$ \checkmark.

Since $\varphi$ and $\psi$ are harmonic (and hence $C^\infty$), all partial derivatives of $p$ and $q$ are continuous. By the Milne-Thomson sufficient conditions:

$\boxed{f = p + iq \text{ is analytic.}}$

Common Traps

• Equality of mixed partials is assumed. The calculation uses $\varphi_{xy} = \varphi_{yx}$ and $\psi_{xy} = \psi_{yx}$; these hold because harmonic functions are $C^\infty$, but state it explicitly if the examiner is strict.
• Each harmonic condition closes exactly one CR equation. $\psi$ harmonic closes $p_x = q_y$; $\varphi$ harmonic closes $p_y = -q_x$. If you mix them up, the terms will not cancel — this is a useful self-check.
• Explicitly invoke the sufficient conditions at the end. State “CR hold and partials are continuous (harmonic functions are smooth), so by the Milne-Thomson theorem $f$ is analytic.” A proof that only shows CR hold without citing continuity of partials is technically incomplete.

Marks-Aware Writing

For a 10-mark CR question: UPSC expects (i) explicit computation of all four partials or limit-definition values, (ii) a clear check of both CR equations with tick marks or “hence CR satisfied”, and (iii) either a path-dependence argument (pathology) or an invocation of the sufficient conditions theorem (analyticity). Showing only one of the two CR equations earns at most half credit.

For the 15-mark recovery question (2020): Show all intermediate steps — both $v_r$ and $v_\theta$, the integration constant argument, and the assembly into a closed form of $z$. The final boxed answer $f(z) = z + 1/z + C$ is worth 3–4 marks; the working leading to it is worth 10–11. Do not skip the $g(\theta)$ determination step.

Theorem statement questions (2023 style): Examiners expect the exact statement of the Milne-Thomson theorem with all three hypotheses: (a) existence of partials, (b) continuity of partials in a neighbourhood, (c) CR equations hold. Omitting (b) is the single most penalised error in this atom.

Practice Set

• 2023-P2-Q3b (15 m) — — analytic function recovery via Milne-Thomson; closely related to the 2020 polar recovery
• 2015-P2-Q1d (10 m) — — verify CR and find $f'$ for an explicit function; good warm-up for the verify-cr archetype
• 2022-P2-Q1b (10 m) — — analytic-constant variant with a different constraint on $|f|$
• 2018-P2-Q1c (10 m) — — CR pathology variant; practise the limit-definition method
• 2017-P2-Q3b (15 m) — — polar recovery with a different harmonic part; longer algebra
• 2016-P2-Q1d (10 m) — — standard analyticity-conditions question; practise theorem statement + log-z application