MatrixTech

A function $f:\mathbb{C}\to\mathbb{C}$ is a map $\mathbb{R}^2\to\mathbb{R}^2$ in disguise — yet complex differentiability is dramatically stronger than the real notion. The reason is one innocent detail: the difference quotient is divided by a complex $h$ , so the limit must come out the same no matter from which direction $h$ shrinks to zero. That single requirement forces the Cauchy–Riemann equations, and out of them fall holomorphy, harmonicity and conformality.

The difference quotient must not care about direction

Definition

Let

U\subseteq\mathbb{C}

be open and

z_0\in U

. Then

f

is complex differentiable at $z_0$ if

f'(z_0)=\lim_{h\to0}\frac{f(z_0+h)-f(z_0)}{h}

with

h\in\mathbb{C}\setminus\{0\}

exists. If this holds at every point of

U

f

is holomorphic on

U

In $\mathbb{R}$ there are two ways to approach a point; in $\mathbb{C}$ there are infinitely many, and all must yield the same complex number. Equivalently, the increment is approximated by multiplication with one fixed complex number, $f(z_0+h)=f(z_0)+f'(z_0)h+o(|h|)$ . Multiplication by $c=a+ib\ne0$ is a rotation by $\arg c$ together with a scaling by $|c|$ — never a shear, never a reflection (for $c=0$ it is the zero map). That is exactly the constraint the real Jacobian will inherit.

Deriving the Cauchy–Riemann equations

Write $z=x+iy$ and split $f(z)=u(x,y)+i\,v(x,y)$ with $u,v:\mathbb{R}^2\to\mathbb{R}$ . Assume $f'(z_0)$ exists; then the limit may be taken along two special directions. Horizontally, $h=t\in\mathbb{R}$ , $t\to0$ : $f'(z_0)=\lim_{t\to0}\frac{[u(x_0+t,y_0)-u]+i\,[v(x_0+t,y_0)-v]}{t}=u_x+i\,v_x.$ Vertically, $h=it$ with $t\to0$ , where the factor $1/i=-i$ appears: $f'(z_0)=\lim_{t\to0}\frac{[u(x_0,y_0+t)-u]+i\,[v(x_0,y_0+t)-v]}{it}=-i\,(u_y+i\,v_y)=v_y-i\,u_y.$ Both are the same complex number, so real and imaginary parts must match separately.

Cauchy–Riemann equations

f=u+iv

is complex differentiable at

z_0

, then the four partials exist there and satisfy

u_x=v_y

and

u_y=-v_x

; moreover

f'(z_0)=u_x+i\,v_x=v_y-i\,u_y

Through the Jacobian this becomes transparent: real differentiability approximates $f$ by $Df=\begin{pmatrix}u_x&u_y\\ v_x&v_y\end{pmatrix}$ , and CR says exactly that this matrix has the shape $\begin{pmatrix}a&-b\\ b&a\end{pmatrix}$ — a rotation–scaling, i.e. multiplication by $a+ib=f'(z_0)$ . In particular $\det Df=u_xv_y-u_yv_x=a^2+b^2=|f'(z_0)|^2\ge0$ : holomorphic maps never reverse orientation.

Is CR enough? Sufficient conditions and one trap

CR is necessary. The converse needs a hypothesis on how the partials behave — four numbers satisfying two equations at a single point is far too weak.

Sufficient criterion

u,v

are real differentiable at

z_0

and CR holds at

z_0

, then

f=u+iv

is complex differentiable at

z_0

. Practical version: if

u_x,u_y,v_x,v_y

exist, are continuous on a neighbourhood of

z_0

and satisfy CR there, then

f

is holomorphic on that neighbourhood.

Pitfall

Take

f(z)=\bar z^{\,2}/z

for

z\ne0

f(0)=0

. Along the axes

u(x,0)=x

v(x,0)=0

u(0,y)=0

v(0,y)=y

, so

u_x=v_y=1

and

u_y=v_x=0

at the origin: CR holds, and

f

is continuous since

|f(z)|=|z|

. Yet for

z=re^{i\theta}

the quotient is

\frac{f(z)-f(0)}{z}=\frac{\bar z^{\,2}}{z^{2}}=e^{-4i\theta}

, which depends on the direction — so

f

is not differentiable at

0

. Real differentiability is the missing ingredient (the partials are discontinuous there). Only the deep Looman–Menchoff theorem rescues a converse:

f

continuous plus CR everywhere on a domain does imply holomorphy.

The Wirtinger form: no dependence on $\bar z$

Treat $z$ and $\bar z$ as formally independent coordinates through the Wirtinger operators $\frac{\partial}{\partial z}=\frac12\Big(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\Big),\qquad \frac{\partial}{\partial\bar z}=\frac12\Big(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\Big).$ Applying the second to $f=u+iv$ gives $\partial_{\bar z}f=\tfrac12[(u_x-v_y)+i\,(v_x+u_y)]$ , which vanishes iff both CR equations hold.

Definition

For real differentiable

f

: complex differentiability at

z_0

\iff

\dfrac{\partial f}{\partial\bar z}(z_0)=0

, and in that case

f'(z_0)=\dfrac{\partial f}{\partial z}(z_0)

This is the fastest test in practice: express $f$ through $z$ and $\bar z$ and hunt for a surviving $\bar z$ . Thus $z^2,\ e^{z},\ \sin z$ are holomorphic, while $\bar z,\ |z|^2=z\bar z,\ \operatorname{Re}z=\tfrac12(z+\bar z)$ are not (at best at isolated points).

Polar form, harmonic pairs, conformality

In polar coordinates $z=re^{i\theta}$ , $r>0$ , the chain rule turns CR into $u_r=\frac1r\,v_\theta,\qquad v_r=-\frac1r\,u_\theta,\qquad f'(z)=e^{-i\theta}(u_r+i\,v_r),$ the convenient form for $\log z$ , $z^{\alpha}$ and anything given by modulus and argument. Since holomorphic functions turn out to be $C^\infty$ (Cauchy's integral formula, later), CR may be differentiated again: $u_{xx}+u_{yy}=(v_y)_x-(v_x)_y=0$ , and likewise for $v$ .

Definition

u

and

v

solve Laplace's equation

\Delta u=0

: they are harmonic, and

v

is a harmonic conjugate of

u

. On a simply connected domain every harmonic

u

has such a conjugate, unique up to a real constant, recovered by integrating

v_x=-u_y

v_y=u_x

. Since

\nabla v=(-u_y,u_x)

\nabla u

rotated by

90^\circ

, the level curves

u=\text{const}

and

v=\text{const}

meet orthogonally wherever

f'(z)\ne0

; at a zero of

f'

of order

m

both gradients vanish and the level sets cross at angle

\pi/(m+1)

instead.

Near a point with $f'(z_0)\ne0$ we have $f(z)\approx f(z_0)+f'(z_0)(z-z_0)$ : rotation by $\arg f'(z_0)$ and uniform stretching by $|f'(z_0)|$ in every direction, so angles between curves survive in size and orientation — $f$ is conformal at $z_0$ . If $f'$ has a zero of order $m$ there, then $f(z)-f(z_0)\sim c\,(z-z_0)^{m+1}$ and angles get multiplied by $m+1$ : under $z\mapsto z^2$ right angles at the origin open into straight angles, so conformality fails exactly at the critical points.

Worked example: testing two functions for holomorphy

Task. Decide where $f(z)=z^2$ and $g(z)=|z|^2$ are complex differentiable, and give the derivative.

Step 1 — Split into real and imaginary parts. $z^2=x^2-y^2+2ixy$ , so $u=x^2-y^2$ , $v=2xy$ ; for $g$ , $u=x^2+y^2$ and $v=0$ .

Step 2 — Partials of $f$ and the CR test. $u_x=2x$ , $u_y=-2y$ , $v_x=2y$ , $v_y=2x$ , hence $u_x=2x=v_y$ ✓ and $u_y=-2y=-v_x$ ✓ for every $(x,y)\in\mathbb{R}^2$ .

Step 3 — Sufficient criterion. These partials are polynomials, hence continuous on all of $\mathbb{C}$ ; together with CR that yields holomorphy, and $f'(z)=u_x+i\,v_x=2x+2iy=2z$ .

Step 4 — Same test for $g$ . $u_x=2x=v_y=0$ forces $x=0$ , and $u_y=2y=-v_x=0$ forces $y=0$ . CR holds only at $z=0$ , so $g$ is differentiable at the origin alone (with $g'(0)=0$ ) and holomorphic nowhere — holomorphy needs an open neighbourhood.

Step 5 — Wirtinger cross-check. $\partial_{\bar z}(z^2)=0$ , while $\partial_{\bar z}(z\bar z)=z$ , which vanishes only at $z=0$ : the same verdict in two lines of work.

Result

f(z)=z^2

is entire with

f'(z)=2z

;

g(z)=|z|^2

is differentiable only at

z=0

and holomorphic on no open set. A single point of differentiability buys you nothing.

Summary

Complex differentiability demands a direction-independent difference quotient — far stronger than the real notion.
Necessarily $u_x=v_y$ and $u_y=-v_x$ ; then $f'=u_x+iv_x$ and $\det Df=|f'|^2\ge0$ .
The converse holds with real differentiability (practically: continuous partials); CR at an isolated point proves nothing.
Compactly $\partial f/\partial\bar z=0$ : a holomorphic function does not see $\bar z$ ; consequently $u,v$ are harmonic conjugates with orthogonal level nets, and $f$ is conformal wherever $f'\ne0$ .

Questions & Answers

Why does one limit condition produce two partial differential equations?

Answer

Comparing just two approach directions already gives two expressions for the single number

f'(z_0)

, namely

u_x+iv_x

and

v_y-iu_y

. Equating them is one complex equation, and a complex equation is two real ones: real parts give

u_x=v_y

, imaginary parts give

v_x=-u_y

. Once

(u,v)

is real differentiable at

z_0

, these two equations force every other direction to give the same limit.

Is complex differentiability the same as real differentiability of $(u,v)$ ?

Answer

No. Real differentiability only asks that

Df

be some linear map; complex differentiability asks it to be

\mathbb{C}

-linear, i.e. of the shape

\begin{pmatrix}a&-b\\b&a\end{pmatrix}

— two extra real conditions on four entries, which is exactly CR. Real differentiability plus CR, however, is equivalent to complex differentiability.

Why does every textbook version of the sufficient condition mention continuous partials?

Answer

Continuity of the partials on a neighbourhood is the standard criterion guaranteeing real differentiability of

u

and

v

: the mean value theorem then produces the linear approximation with an

o(|h|)

error. It is not the weakest hypothesis: real differentiability is weaker, and Looman–Menchoff needs even less — but only as a statement on a whole domain (

f

continuous, all four partials existing everywhere, CR everywhere). Continuity of the partials, however, is the one you can verify at a glance.

What exactly fails in the counterexample $f(z)=\bar z^{\,2}/z$ ?

Answer

All four partials exist at the origin and satisfy CR, and

f

is continuous there. But the difference quotient equals

e^{-4i\theta}

along the ray of angle

\theta

, so it sweeps the entire unit circle as the direction varies. In fact

f

is not real differentiable at

0

— the difference quotient admits no linear approximation at all; consistently, the partials turn out to be discontinuous there.

What is the difference between "differentiable at $z_0$ " and "holomorphic at $z_0$ "?

Answer

Holomorphic at

z_0

means complex differentiable on an entire open neighbourhood, not merely at the point.

|z|^2

shows the distinction has teeth: differentiable at

0

, holomorphic nowhere. Essentially every strong theorem — analyticity, Cauchy's theorem, the maximum principle — needs the open-set version.

Why is $\partial f/\partial\bar z=0$ such a useful reformulation?

Answer

It converts a check on four partials into pattern recognition: rewrite

f

in terms of

z

and

\bar z

, and any surviving

\bar z

blocks holomorphy wherever its coefficient is non-zero. It also makes the slogan "holomorphic = independent of the conjugate variable" literal and generalises directly to several variables and to the

\bar\partial

-problems of complex geometry.

Does every harmonic function come from a holomorphic one?

Answer

Locally yes: on a disc every harmonic

u

equals

\operatorname{Re}f

for some holomorphic

f

, obtained by integrating

v_x=-u_y

v_y=u_x

. Globally the topology matters — on the punctured plane

u=\log|z|

is harmonic but its conjugate is

\arg z

, which admits no single-valued branch. Simple connectedness is exactly the hypothesis that removes the obstruction.

Do the CR equations have a physical reading?

Answer

Yes. If

f=u+iv

is holomorphic, then

\nabla u=\overline{f'(z)}

is simultaneously divergence-free (because

\Delta u=0

) and curl-free (being a gradient):

f

is the complex potential of an ideal planar flow with velocity potential

u

and stream function

v

, whose streamlines are the curves

v=\text{const}

. The same pair models electrostatic potentials and steady temperature fields, which is why conformal mapping became a classical tool for solving Laplace's equation on awkward domains.

The rest of the map

Complex differentiability is the entrance gate; everything downstream in complex analysis is a consequence of the rigidity you have just met.

Integration theory

Cauchy's integral theorem — contour integrals of holomorphic functions vanish on simply connected domains: the integral counterpart of CR.
Cauchy's integral formula — interior values are fixed by boundary values; the source of infinite differentiability.
Goursat's theorem — Cauchy's theorem without assuming $f'$ continuous, the technically sharpest entry point.

Local structure and global consequences

Power series and analyticity — holomorphic $\Rightarrow$ locally a convergent Taylor series, so holomorphic and analytic coincide.
Identity theorem — agreement on a set with an accumulation point forces global agreement: rigidity at its most extreme.
Laurent series, singularities, residues — what happens where holomorphy fails at isolated points, and how to exploit it.
Liouville and maximum modulus — bounded entire functions are constant; $|f|$ attains no interior maximum.

Geometry and applications

Riemann mapping theorem — every simply connected proper subdomain of $\mathbb{C}$ is conformally equivalent to the unit disc.
Möbius transformations — the conformal automorphisms of the sphere and the working examples for all of the above.
Potential theory — Dirichlet problems solved by transporting harmonic functions along conformal maps.

Suggested roadmap

Master the CR test in both forms (partials and $\partial_{\bar z}$ ) on $z^n$ , $e^z$ , $\bar z$ , $|z|^2$ , $\log z$ .
Compute harmonic conjugates on discs and locate the obstruction on an annulus.
Prove Goursat's lemma, then Cauchy's theorem for convex domains.
Derive the Cauchy integral formula and deduce analyticity and Liouville.
Develop Laurent series, classify isolated singularities, and learn the residue theorem.
Return to geometry: Möbius maps, conformal equivalence, the Riemann mapping theorem.

Everything on that list is powered by the observation you started with: one limit that refuses to depend on direction.

The difference quotient must not care about direction

Deriving the Cauchy–Riemann equations

Is CR enough? Sufficient conditions and one trap

The Wirtinger form: no dependence on zˉ\bar zzˉ

Polar form, harmonic pairs, conformality

Worked example: testing two functions for holomorphy

Summary

Questions & Answers

The rest of the map

The Wirtinger form: no dependence on $\bar z$