Chapter 4 · Building analytic functions: harmonic conjugates

The hook

Given the real part, find the function

Story

In Chapter 3 you learned to test for analyticity. Given $f(z) = u(x,y) + i\,v(x,y)$ , the Cauchy-Riemann equations decide whether it is analytic, and as a free consequence the real part $u$ and the imaginary part $v$ are both harmonic — each one solves Laplace's equation $u_{xx} + u_{yy} = 0$ .

Now flip the question. Suppose somebody hands you only the real part. They write $u(x, y) = x^2 - y^2$ and ask: is there an analytic function whose real part is exactly this? If yes, what is the imaginary part $v$ ? And what is the function $f(z)$ as an expression in $z$ ?

Three lines of work and the answer is $f(z) = z^2$ . The imaginary part you needed was $v = 2xy$ , and the function you reconstructed is the simplest non-trivial polynomial in $z$ . This chapter teaches the two recipes that produce that answer — and choose the right one under exam pressure.

Why this matters

The harmonic-conjugate construction has a real-world payoff. Harmonic functions on the plane are the natural model for steady-state heat distribution, ideal fluid flow, and electrostatic potential. The construction in this chapter says: every such physics problem in the plane can be lifted to a problem about an analytic function, where the machinery of complex analysis becomes available. That lift is what conformal mapping (Chapter 5) exploits.

For now, you don't need to chase any of that. You need to learn the construction well enough to do it under exam pressure on five fresh $u$ 's in twenty minutes.

The setup

The contract:

The harmonic-conjugate problem

\text{Given: a harmonic } u(x, y). \qquad \text{Find: } v(x, y) \text{ such that } f(z) = u + iv \text{ is analytic.}

Such a $v$ is called a harmonic conjugate of $u$ . Conjugate here does not mean complex conjugate — it means partner under CR. The pair $(u, v)$ is conjugate when together they make an analytic function.

Two facts used freely after this beat.

Existence. On simply connected domains — the ones you will work with, typically all of $\mathbb{R}^2$ — every harmonic $u$ admits a conjugate $v$ .
Uniqueness. On any connected domain, $v$ is unique up to an additive real constant. (Quick proof: if $v_1$ and $v_2$ are both conjugates of $u$ , the difference $v_1 - v_2$ has both partials zero by CR, hence is constant on a connected set.) So every answer ends with a free real parameter that an extra condition like $f(0) = 0$ pins down.

The symmetric problem

If you are instead given $v(x, y)$ and asked for $u$ , the same machinery works with CR signs flipped. Milne-Thomson becomes $f'(z) = v_y(z, 0) + i\,v_x(z, 0)$ — same recipe, swapped partials, plus sign instead of minus. The exam translator in Beat 6 walks through this variant.

Intuition before formula

Why harmonicity is exactly the condition

The CR equations from Chapter 3, rearranged with $u$ known and $v$ unknown:

CR, read as a system for

v

v_x = -u_y, \qquad v_y = u_x

This tells you both partial derivatives of $v$ in terms of the given function $u$ . If you know both partials, you can recover $v$ by integration — provided the two expressions are consistent with each other.

The consistency check is Clairaut's theorem (mixed partials commute). Differentiating the first equation with respect to $y$ and the second with respect to $x$ :

\frac{\partial}{\partial y}(v_x) = -u_{yy}, \qquad \frac{\partial}{\partial x}(v_y) = u_{xx}

For these to agree, $u_{xx} + u_{yy} = 0$ — exactly Laplace's equation.

Insight

This is genuinely elegant. The Chapter 3 fact "if $f$ is analytic then $u$ is harmonic" looked like a free bonus. Now you see it from the other side: harmonicity of $u$ is not just a consequence of analyticity — it is the condition under which $u$ can be the real part of an analytic function in the first place.

So whenever a question begins "given a harmonic function $u = \dots$ ," the harmonicity is not flavour text. It is the green light that says "yes, the conjugate you are about to find exists."

One technical note. Harmonicity is a local condition — it pins down $v$ up to a constant in any small neighbourhood. Getting from there to a single globally defined $v$ needs the domain to be simply connected, which is why that hypothesis sits inside the existence statement. For this chapter, every domain you meet is $\mathbb{R}^2$ or an open disc, so the global step is automatic.

The two methods

Direct method, Milne-Thomson, and the orthogonal-families bonus

Method 1 — Direct method

The recipe.

Step 1. Compute $u_x$ and $u_y$ . Verify $u_{xx} + u_{yy} = 0$ (harmonic check).
Step 2. From CR: $v_x = -u_y$ . Integrate with respect to $x$ , treating $y$ as constant: $v = \int (-u_y)\,dx + g(y)$ . The constant of integration is a function $g(y)$ of $y$ alone.
Step 3. Differentiate this $v$ with respect to $y$ and set the result equal to $u_x$ (the second CR equation). Solve for $g'(y)$ . Integrate to get $g(y)$ .
Step 4. Assemble $v$ , then $f(z) = u + iv$ . Re-express in $z$ by substituting $x = (z + \bar z)/2$ , $y = (z - \bar z)/(2i)$ and watching the $\bar z$ terms cancel.

Worked example. $u = x^2 - y^2$ . Then $u_x = 2x$ , $u_y = -2y$ , and $u_{xx} + u_{yy} = 2 - 2 = 0$ ✓. So $v_x = 2y$ , giving $v = 2xy + g(y)$ . Then $v_y = 2x + g'(y) = u_x = 2x$ , so $g'(y) = 0$ and $g = C$ with $C \in \mathbb{R}$ . Therefore $v = 2xy + C$ and

f(z) = (x^2 - y^2) + i(2xy + C) = (x + iy)^2 + iC = z^2 + iC.

The free constant is pure imaginary — it lives entirely in $v$ . Adding a real constant to $f$ would shift the real part by that real number and stop matching the prescribed $u$ .

Method 2 — Milne-Thomson

The shortcut. This is the method to use on exams unless the question explicitly asks for $v(x, y)$ without asking for $f(z)$ .

Step 1. Compute $\varphi_1(x, y) = u_x$ and $\varphi_2(x, y) = u_y$ .
Step 2. Substitute $x \to z$ and $y \to 0$ in both: $\varphi_1(z, 0)$ and $\varphi_2(z, 0)$ .
Step 3. Then $f'(z) = \varphi_1(z, 0) - i\,\varphi_2(z, 0)$ .
Step 4. Integrate with respect to $z$ . The result is $f(z) = F(z) + iC$ with $C \in \mathbb{R}$ . An extra condition like $f(0) = 0$ or $f(1) = 2 + i$ pins down $C$ .

Worked example. Same $u = x^2 - y^2$ . Then $\varphi_1 = 2x$ , so $\varphi_1(z, 0) = 2z$ . And $\varphi_2 = -2y$ , so $\varphi_2(z, 0) = 0$ . So $f'(z) = 2z - i \cdot 0 = 2z$ and $f(z) = z^2 + iC$ . Two lines. Same answer as the direct method.

Where the recipe comes from

If $f$ is analytic, then $f'(z) = u_x + i\,v_x = u_x - i\,u_y$ (the second equality is CR). So $f'(z)$ , viewed as a function of $x$ and $y$ , is built entirely from partials of $u$ . The substitution $x \to z, y \to 0$ then produces an expression in $z$ alone, which we integrate. Why the substitution is legitimate uses the fact that an analytic function on a connected domain is fully determined by its values on a real interval — a result we will meet in Chapter 9, when we study power-series representations of analytic functions. For this chapter, treat Milne-Thomson as a reliable construction rule and always sanity-check the final $f(z)$ by computing $\operatorname{Re}(f(z))$ and confirming it matches the given $u$ .

The orthogonal-families picture

Here is a free geometric fact that often shows up on exams. Fix an analytic $f = u + iv$ . Consider the two families of level curves:

u(x, y) = c_1 \quad (\text{curves on which the real part is constant})

v(x, y) = c_2 \quad (\text{curves on which the imaginary part is constant})

These two families cross each other at right angles at every point where $f'(z) \neq 0$ .

The reason is one line of vectors. At any point where $\nabla u \neq 0$ , the gradient $\nabla u = (u_x, u_y)$ is perpendicular to the level curve $u = c_1$ through that point. Similarly $\nabla v$ is perpendicular to the curve $v = c_2$ . Now compute the dot product, using CR ( $v_x = -u_y,\ v_y = u_x$ ):

\nabla u \cdot \nabla v = u_x v_x + u_y v_y = u_x(-u_y) + u_y(u_x) = 0.

The gradients are perpendicular, and rotating each gradient by $90°$ gives the tangent direction of the corresponding level curve — so the tangents are also perpendicular. Where $f'(z) = u_x - i\,u_y = 0$ , both partials of $u$ vanish, $\nabla u = 0$ , and the level curve has no well-defined normal direction; orthogonality can fail there.

Insight

For $f(z) = z^2$ : $u = x^2 - y^2$ is a family of hyperbolas, and $v = 2xy$ is another family of hyperbolas rotated by $45°$ . They cross at right angles everywhere except at the origin, where $f'(0) = 0$ . You will see this in the playground next — and the same orthogonality for three other analytic functions.

This shows up on exams as: "Show that the families $u = c_1$ and $v = c_2$ cut each other orthogonally." Five marks, three lines of work using CR.

Interactive playground

Pick a $u$ , watch both methods produce the same $f(z)$

Make the speed difference visible. On the same $u$ , side by side, you will see both methods produce the same $f(z)$ and one of them takes noticeably fewer steps. The harmonic check is pulled out as Step 0 — it is a property of $u$ , not of either method, and the two methods are interchangeable downstream. The geometric payoff — orthogonal level curves — appears below both columns once they have finished, attributed to $f$ itself rather than to either construction route.

Harmonic-conjugate race · Direct vs Milne-Thomson

u₁ = x² − y² (Re z²) u₂ = x³ − 3xy² (Re z³) u₃ = eˣ cos y (Re eᶻ) u₄ = sinh x · cos y (Re sinh z)

Step 0 · Harmonic check (one-time, method-independent)

Direct method

Milne-Thomson

The geometric picture of f(z)

u = c₁ v = c₂

Insight

A subtle thing the playground reveals on $u_3 = e^x \cos y$ . Under Milne-Thomson the substitution $y = 0$ turns $e^x \cos y$ into $e^z \cdot 1 = e^z$ and $-e^x \sin y$ into $-e^z \cdot 0 = 0$ . The $f'(z) = e^z - i \cdot 0 = e^z$ falls out instantly, and $f(z) = e^z + iC$ . The direct method on the same $u$ takes a few extra lines of $g(y)$ chasing to reach the same answer. The difference is not "Milne-Thomson is shorter" — it is "Milne-Thomson skips integrating in two variables and integrates in one."

Retrieval gauntlet

Test yourself before moving on

Do not peek. Commit to an answer before you reveal anything. Recognising a correct answer once you see it only feels like learning.

Quiz 1 · spot the non-harmonic

Which of the following cannot be the real part of an analytic function on $\mathbb{R}^2$ ?

Quiz 2 · Milne-Thomson, sign check

Given $u = e^x \cos y$ , what is $f'(z)$ by Milne-Thomson?

Quiz 3 · orthogonal families

For $f(z) = z^2$ , the curve $u = 1$ is the hyperbola $x^2 - y^2 = 1$ . The curve $v = 1$ is the hyperbola $2xy = 1$ . At a point where the two hyperbolas cross, the angle between them is:

Reveal 1 — Direct method, scaffolded. Given $u = x^3 - 3xy^2$ , find the conjugate $v$ and express $f(z)$ . Work the three sub-steps below; attempt each on paper before revealing it.

Step 1. Check that

u

is harmonic, and write down the two CR targets

v_x

and

v_y

in terms of

u

.

$u_x = 3x^2 - 3y^2$ . $u_y = -6xy$ . $u_{xx} = 6x$ , $u_{yy} = -6x$ . Sum is $0$ ✓.

Targets: $v_x = -u_y = 6xy$ ; $v_y = u_x = 3x^2 - 3y^2$ .

Step 2. Integrate

v_x

with respect to

x

to get

v

up to a function of

y

.

$v = \displaystyle\int 6xy\,dx = 3x^2 y + g(y)$ .

Step 3. Pin down

g(y)

using the second CR equation

v_y = u_x

, and write

f(z)

.

$v_y = 3x^2 + g'(y)$ . Set equal to $3x^2 - 3y^2$ . So $g'(y) = -3y^2$ , hence $g(y) = -y^3 + C$ with $C \in \mathbb{R}$ . Therefore $v = 3x^2 y - y^3 + C$ and

f(z) = (x^3 - 3xy^2) + i(3x^2 y - y^3 + C) = (x + iy)^3 + iC = z^3 + iC.

The free constant rides on $v$ ; it is pure imaginary in $f$ .

Reveal 2 — Milne-Thomson and the $y = 0$ trap. Given $u = \sinh x \cos y$ , find $f(z)$ by Milne-Thomson.

Step 1. Compute

\varphi_1 = u_x

and

\varphi_2 = u_y

.

$\varphi_1 = \cosh x \cos y$ . $\varphi_2 = -\sinh x \sin y$ .

Step 2. Substitute

x \to z

AND

y \to 0

. (This is the step students botch.)

$\varphi_1(z, 0) = \cosh z \cdot \cos 0 = \cosh z \cdot 1 = \cosh z$ .

$\varphi_2(z, 0) = -\sinh z \cdot \sin 0 = -\sinh z \cdot 0 = 0$ .

The trap: writing $\varphi_1(z, 0) = \cosh z \cos z$ — i.e., renaming $y$ as $z$ instead of zeroing it out. The $y = 0$ substitution is what collapses the two-variable function back to a one-variable function of $z$ . Always write the substitution as a visible step on your page.

Step 3. Assemble

f'(z)

, integrate, and sanity-check by reading off

\operatorname{Re}(f)

.

$f'(z) = \cosh z - i \cdot 0 = \cosh z$ . Then $f(z) = \displaystyle\int \cosh z\,dz = \sinh z + iC$ with $C \in \mathbb{R}$ .

Check: $\operatorname{Re}\bigl(\sinh(x + iy)\bigr) = \sinh x \cos y$ . ✓ Matches the given $u$ .

Exam translator

How this concept appears on exams

Exam trap

In standard Indian engineering-math exams — VTU 21MAT41, JIIT Maths-II and their cousins — and likely in MCC201A unless your class notes differ, this chapter shows up in three forms.

CO-3 form, main pattern. "Given $u(x, y) = \dots$ , show that $u$ is harmonic and find the analytic function $f(z) = u + iv$ . Express your answer in terms of $z$ ." Roughly 6–8 marks, budget 8 minutes. Use Milne-Thomson. Direct integration takes 12–15 minutes on a transcendental $u$ and increases sign-error risk.

CO-3 form, given the imaginary part. "Given $v(x, y) = \dots$ , find the analytic function $f(z)$ ." Same machinery, CR signs flipped. From $f' = u_x + i\,v_x$ and CR ( $u_x = v_y,\ u_y = -v_x$ ), Milne-Thomson becomes

$f'(z) = v_y(z, 0) + i\,v_x(z, 0)$

Note the plus sign (not minus) and the swapped partials. Integrate to get $f(z) = F(z) + C$ with $C \in \mathbb{R}$ — here the free constant is pure real, because $v$ is fixed and the freedom lives in $u$ . Mirror image of the $u$ -given case, where the free constant was pure imaginary.

CO-3 form, orthogonal families variant. "Show that the families of curves $u = c_1$ and $v = c_2$ cut each other orthogonally." Roughly 4–5 marks, budget 5 minutes. Compute $\nabla u \cdot \nabla v$ and use CR; the dot product reduces to zero.

Occasionally the question pins the constant: "Find $f(z)$ such that $f(0) = 0$ " or similar. Your answer is $f(z) = F(z) + iC$ with $C \in \mathbb{R}$ . Plug the given point into this expression and solve for $C$ . Two marks, ten seconds — often skipped under exam pressure. Don't skip it.

Four patterns that cost marks

Stopping at $v(x, y)$ and not expressing the answer in $z$ . The single most common mark-loss on these questions. You compute $v$ correctly, write $f(z) = u + iv$ with $x$ and $y$ still in it, and lose 2–3 marks on the "express in terms of $z$ " step the question demanded. The fix: substitute $x = (z + \bar z)/2$ , $y = (z - \bar z)/(2i)$ and watch the $\bar z$ 's cancel — or, with Milne-Thomson, you arrive in $z$ form directly. If you used the direct method, the last line of working must be $f(z) = \text{(something in } z\text{)} + iC$ . Never stop earlier.
Forgetting the $y = 0$ substitution. The biggest Milne-Thomson trap. Write the substitution as a visible step on the page — actually write "let $x = z$ , $y = 0$ " — before evaluating. A second of effort, several marks.
Sign error on the $-i\,\varphi_2$ term. The formula has a minus: $f'(z) = \varphi_1(z, 0) - i\,\varphi_2(z, 0)$ . Recall it via the derivation: $f' = u_x + i\,v_x = u_x - i\,u_y$ (the second equality uses CR). The minus is not arbitrary — it is the CR minus sign you already memorised. (And for the symmetric $v$ -given form, the sign flips to plus and the partials swap.)
Skipping the harmonic check on a question that demands it. If the question says "show that $u$ is harmonic and find …," the marks for the harmonic check are explicitly there. Compute $u_{xx} + u_{yy}$ and write $= 0$ . Two marks for ten seconds of work.

Five things to carry out of this chapter

The harmonic-conjugate problem is the inverse of CR. Given the real part $u$ of an analytic function, you can recover the imaginary part $v$ (up to an additive real constant) and hence $f(z) = F(z) + iC$ with $C \in \mathbb{R}$ .
Harmonicity of $u$ is the local condition that makes the construction work. On the simply connected domains you work with, that local condition is enough for global existence.
The direct method. $v_x = -u_y$ , integrate in $x$ → $v = (\text{piece}) + g(y)$ ; then $v_y = u_x$ pins $g'(y)$ ; combine.
Milne-Thomson. $\varphi_1 = u_x$ , $\varphi_2 = u_y$ ; substitute $x \to z$ and $y \to 0$ ; then $f'(z) = \varphi_1(z, 0) - i\,\varphi_2(z, 0)$ ; integrate. Faster, exam-preferred.
Orthogonal families. The level curves of $u$ and of $v$ cross at right angles everywhere $f'(z) \neq 0$ . One-line proof via $\nabla u \cdot \nabla v = 0$ from CR.

Why this matters next

A note on the syllabus phrase "elementary analytic functions." The MCC201A Unit 1 contents list "elementary analytic functions" as one of its topics. Those functions are $e^z, \sin z, \cos z, \sinh z, \cosh z$ — and you have just used several of them as the worked examples in this chapter. You already know what they look like, decomposed into real and imaginary parts. The logarithm $\log z$ is also "elementary," but it carries a branch-cut subtlety that needs its own deliberate treatment; you'll meet it where it does real work, in the contour-integration chapter.

What comes next. Chapter 5, Conformal mapping intuition, takes the analytic functions you have learned to recognise (Chapter 3) and construct (Chapter 4), and asks what they do geometrically. The orthogonal-families picture from Beat 3 is the first hint — analytic functions reshape regions of the plane while preserving angles. The functions $z^2$ , $e^z$ , $\sinh z$ will reappear immediately as conformal maps with specific geometric effects.

Before you open Chapter 5: solve five fresh "given $u$ , find $f(z)$ " problems from Zill, Complex Analysis, Chapter 3 (harmonic-functions section) [verify the exact section number against your copy]. Mix two by direct method, three by Milne-Thomson. 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 5 — Harmonic Functions — harmonic functions and conjugates, the same material worked a different way.

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

Given the real part, find the function

The setup

Why harmonicity is exactly the condition

Direct method, Milne-Thomson, and the orthogonal-families bonus

Method 1 — Direct method

Method 2 — Milne-Thomson

The orthogonal-families picture

Pick a uuu, watch both methods produce the same f(z)f(z)f(z)

Test yourself before moving on

How this concept appears on exams

Four patterns that cost marks

Five things to carry out of this chapter

Same material, another voice

Pick a $u$ , watch both methods produce the same $f(z)$