Chapter 3 · The Cauchy-Riemann detective test

The hook

The setup

Story

Real calculus is like driving on a one-lane road. Forward, back. $f(x)$ takes a number and gives a number. Easy.

Complex calculus is like flying a drone. You can approach any point from infinitely many directions — north, southwest, spiral, zigzag. $f(z)$ takes a point in a plane and gives a point in a plane.

Here is the kicker: for a function to be differentiable in this 2D world, the answer must be the same no matter which direction you approach from.

You met that idea in Chapter 2, in the direction-of-approach playground. Now we put it to work. That one demand — same answer from every direction — is an absurdly strict standard. Most functions fail it instantly, at almost every point. The few that pass are a rare and powerful species.

If real differentiability is "your friend is reliable on Mondays," complex differentiability is "your friend is reliable on Mondays, Tuesdays, weekends, leap days, and from every direction approaching every moment from infinity." That is a hard friend to find — and that is exactly why the functions that pass get their own name. They are called analytic, and that rarity is the whole reason they are so useful.

Why this matters

Analytic functions describe fluid flow, electric fields, signal processing, and quantum mechanics. The constraint that makes them rare is precisely what makes them useful. And in your exams: every contour integral, every residue, every conformal map in Unit 1 assumes analyticity. Get this test wrong and the rest of the unit collapses under you. This chapter is the foundation — so we build it carefully.

Intuition before formula

The two-rule test

We will not hand you the Cauchy-Riemann equations as a magic spell. We will derive them, so they feel inevitable.

Write a complex function by splitting it into a real part and an imaginary part: $f(z) = u(x,y) + i\,v(x,y)$ . The derivative $f'(z_0)$ is defined as a limit — and because $z$ is a point in a plane, that limit can be approached in many ways. Demand agreement on just two of them — along the real axis and along the imaginary axis — and the constraint already bites hard.

Approach 1 — along the real axis. Hold $y$ fixed and let $\Delta x \to 0$ :

f'(z_0) = \lim_{\Delta x \to 0} \frac{u(x+\Delta x,\,y) + i\,v(x+\Delta x,\,y) - u(x,y) - i\,v(x,y)}{\Delta x} = \frac{\partial u}{\partial x} + i\,\frac{\partial v}{\partial x}

Approach 2 — along the imaginary axis. Hold $x$ fixed and let $\Delta y \to 0$ . The step in $z$ is now $i\,\Delta y$ , so the denominator carries that $i$ :

f'(z_0) = \lim_{\Delta y \to 0} \frac{u(x,\,y+\Delta y) + i\,v(x,\,y+\Delta y) - u(x,y) - i\,v(x,y)}{i\,\Delta y} = \frac{\partial v}{\partial y} - i\,\frac{\partial u}{\partial y}

If $f$ is differentiable at $z_0$ , these two results must be the same complex number. Two complex numbers are equal exactly when their real parts match and their imaginary parts match. Equate them:

The Cauchy-Riemann equations

\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \qquad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}

That is the whole test. Two equations. Notice they were not invented — they fell out of one honest demand: the derivative cannot depend on the direction of approach.

Story

Now that you have derived them, here is how to remember them. Picture two bouncers at a club door. Bouncer 1 checks: "Does $u$ 's $x$ -rate match $v$ 's $y$ -rate?" Bouncer 2 checks: "Does $u$ 's $y$ -rate equal the negative of $v$ 's $x$ -rate?" The minus sign is the trap. The large majority of students who lose marks here lost them to that minus.

Insight

The move examiners love: they hand you $u$ and $v$ , ask "is this analytic?", and choose an example where rule 1 passes but rule 2 fails — or the other way round. One passing rule proves nothing. You must check both, every time.

Formal statement

What "analytic" actually requires

A definition and a test — kept deliberately separate, so the logic never runs in a circle.

Definition. A function $f(z)$ is analytic in an open region if it is complex differentiable at every point of that region. Analyticity is a property of regions, not of single points. Differentiability at one isolated point does not earn the word.

The engineering-math test. The Cauchy-Riemann equations on their own are necessary but not sufficient. The safe, sufficient test you will actually use on exam problems is this:

The test to use

If $u(x,y)$ and $v(x,y)$ have continuous first partial derivatives throughout a region, and the Cauchy-Riemann equations hold at every point of that region, then $f = u + iv$ is analytic there.

Two ingredients: continuous partials, and CR everywhere in the region. Name both when you write a solution — the second mark is often for saying "continuous partials" out loud.

Exam trap

A function can be differentiable at a single isolated point and yet analytic nowhere. The textbook example is $f(z) = |z|^2$ , so $u = x^2 + y^2$ and $v = 0$ . Compute the partials: $\partial u/\partial x = 2x$ , $\partial u/\partial y = 2y$ , $\partial v/\partial x = 0$ , $\partial v/\partial y = 0$ . Rule 1 needs $2x = 0$ ; rule 2 needs $2y = 0$ . Both hold only at $z = 0$ . So $f$ is differentiable at the origin — but in any region around it, CR fails at every other point, so $f$ is analytic in no region at all.

If a question says "differentiable at $z = 0$ ," that is not the same as "analytic." Read the wording like a detective.

One more form worth knowing. For functions written naturally in polar coordinates — $\log z$ is the classic case — the Cauchy-Riemann equations take this shape:

Cauchy-Riemann equations, polar form

\frac{\partial u}{\partial r} = \frac{1}{r}\frac{\partial v}{\partial \theta}, \qquad \frac{\partial v}{\partial r} = -\frac{1}{r}\frac{\partial u}{\partial \theta}

Same structure as the Cartesian pair — and the minus sign still rides on the second equation.

Insight

A beautiful consequence. If $f = u + iv$ is analytic, then $u$ and $v$ separately satisfy Laplace's equation, $\partial^2 u/\partial x^2 + \partial^2 u/\partial y^2 = 0$ . Functions that do this are called harmonic. So every analytic function quietly hands you two solutions of Laplace's equation for free — which is why complex analysis turns up everywhere in engineering. You will prove this yourself, step by step, in the next section.

Interactive playground

Try it: the detective tool

Pick a function below. Watch its partial derivatives appear, watch the two rules get checked, and read the verdict. Work through all five — the pattern across them is the real lesson.

Cauchy-Riemann equation checker

Function:

Insight

The pattern. The "natural" functions of $z$ — polynomials, the exponential, sine and cosine — pass. The functions that secretly depend on $\bar{z}$ , the conjugate — like $|z|^2 = z\bar{z}$ , or $\bar{z}$ itself — fail. This is the deep rule of the whole subject: analytic means depends on $z$ , never on $\bar{z}$ . Spot a conjugate hiding in a problem, and you have already smelled the trap.

Retrieval gauntlet

Test yourself before moving on

Do not peek. Commit to an answer before you reveal anything. Committing — even to a wrong answer — is what builds the memory. Recognising a correct answer once you see it only feels like learning.

Quiz 1 · check the test

A function has $\partial u/\partial x = 3$ and $\partial v/\partial y = 3$ . Rule 1 passes. The student writes "analytic" and moves on. What is the problem?

Try first. Show that

u(x,y) = x^2 - y^2

is harmonic. Compute

\partial^2 u/\partial x^2 + \partial^2 u/\partial y^2

on paper before you reveal.

$\partial u/\partial x = 2x$ , so $\partial^2 u/\partial x^2 = 2$ . And $\partial u/\partial y = -2y$ , so $\partial^2 u/\partial y^2 = -2$ . Their sum is $2 + (-2) = 0$ . So $u$ is harmonic.

Notice $u = x^2 - y^2$ is exactly the real part of $z^2 = (x^2 - y^2) + i(2xy)$ . Its partner $v = 2xy$ is the harmonic conjugate — and constructing it from $u$ is the whole job of Chapter 4.

Quiz 2 · spot the trap

Which function is differentiable at $z = 0$ but not analytic in any region?

Prove: if $f = u + iv$ is analytic, then $u$ is harmonic — that is, $u$ satisfies Laplace's equation $\partial^2 u/\partial x^2 + \partial^2 u/\partial y^2 = 0$ . Work the three sub-steps below. Attempt each on paper before revealing it.

Step 1. Take the first Cauchy-Riemann equation,

\partial u/\partial x = \partial v/\partial y

. Differentiate both sides with respect to

x

. What do you get?

\frac{\partial^2 u}{\partial x^2} = \frac{\partial^2 v}{\partial x \, \partial y}

Step 2. Take the second Cauchy-Riemann equation,

\partial u/\partial y = -\partial v/\partial x

. Differentiate both sides with respect to

y

. What do you get?

\frac{\partial^2 u}{\partial y^2} = -\frac{\partial^2 v}{\partial y \, \partial x}

Step 3. Add your two results. What fact from multivariable calculus do you invoke about the right-hand sides — and what do you conclude?

Add step 1 and step 2. The right-hand side becomes $\dfrac{\partial^2 v}{\partial x \, \partial y} - \dfrac{\partial^2 v}{\partial y \, \partial x}$ .

Provided the second partial derivatives are continuous, Clairaut's theorem says the mixed partials are equal: $\dfrac{\partial^2 v}{\partial x \, \partial y} = \dfrac{\partial^2 v}{\partial y \, \partial x}$ . So the right-hand side is $0$ , and therefore

\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0

which is exactly the statement that $u$ is harmonic. ∎

In a rigorous complex-analysis course, the smoothness of $u$ and $v$ follows from analyticity as a theorem. For engineering-math exams that regularity is assumed when proving harmonicity from CR. Memorise the structure of this proof — it appears on essentially every Indian engineering-math paper that touches analytic functions.

Quiz 3 · the polar form

Which of these is the correct second polar-form Cauchy-Riemann equation?

Exam translator

How this concept appears on exams

Exam trap

In standard Indian engineering-math exams, and likely in MCC201A unless your class notes differ, Cauchy-Riemann shows up in two forms within this chapter's territory.

CO-1 / CO-2 form — explain, state, sometimes prove. "State the Cauchy-Riemann equations. State the conditions for analyticity in a region. Show that if $f(z) = u + iv$ is analytic then $u$ and $v$ are harmonic." Roughly 5–7 marks; budget about 8 minutes. The proof is the scaffolded one you just worked through.

CO-3 form — solve a simple problem. "Verify whether $f(z) = [\text{expression}]$ is analytic. If so, state where." Roughly 5–8 marks; budget about 6 minutes. The five functions in this chapter's playground are essentially the menu these are drawn from.

A third form — given $u$ , construct $f(z)$ — also exists, but that is Chapter 4's territory and uses a method we have not met yet.

The three patterns that cost marks.

A sign error on rule 2 — forgetting the minus on $\partial v/\partial x$ , in either the Cartesian or the polar form.
Confusing "differentiable at a point" with "analytic in a region." This quietly costs 2–3 marks on a typical paper.
Not stating that the first partials are continuous when you claim sufficiency. The CR equations alone do not finish the argument.

Six things to carry out of this chapter

Analytic means complex differentiable at every point of an open region. A brutally strict standard — and a rare one.
Cauchy-Riemann is the two-rule test, derived by approaching the derivative along the real and the imaginary axes. Check both rules — including the minus sign.
The engineering-math test: continuous first partials, plus CR holding everywhere in a region, gives analyticity in that region.
Point versus region. "Differentiable at $z_0$ " is not "analytic in a region." Watch for $|z|^2$ and its relatives.
Real and imaginary parts of an analytic function are both harmonic — they satisfy Laplace's equation. The proof leans on Clairaut's theorem for mixed partials.
Functions of $\bar{z}$ are never analytic. Spot the conjugate, smell the trap.

Why this matters

What comes next. Chapter 4, Harmonic Conjugates, runs this machinery in reverse: given a harmonic $u$ , construct the analytic $f(z)$ — by the direct method or the Milne-Thomson shortcut. That is a different mental model, so it gets its own chapter.

Before you open Chapter 4: attempt three fresh CR-verification problems from Zill, Complex Analysis, Chapter 3 (the sections on the Cauchy-Riemann equations) [verify the exact section number against your copy]. Choose problems where the function is given as a single expression in $x$ and $y$ , and decide analyticity. Re-reading this chapter is a trap — retrieval beats recognition every time.

Same material, another voice

If a different explanation would help, this one is worth your time — free, from MIT OpenCourseWare:

Read: Orloff, MIT 18.04, Topic 2 — Analytic Functions — the Cauchy–Riemann equations and what makes a function analytic, the heart of this chapter.

✓ Chapter complete. Your progress, and every quiz answer, is saved on this computer — revisit any time.