Mathematical Analysis · Complex Analysis

Regular Functions

Multiplying by a complex number is a rotational homothety (see «Complex Numbers: Three Constructions of One Field»). Which functions $f(z)$ behave just as well at every point? It turns out to be a rigid and surprisingly narrow family: regular functions, for which mere differentiability implies representability as a power series.

1. The Derivative and Regularity

The complex derivative is defined by the same limit as for functions of one real variable:

$$f'(z)=\lim_{\Delta z\to0}\frac{f(z+\Delta z)-f(z)}{\Delta z},$$

except now the increment $\Delta z$ is a complex number, and it can tend to zero from any direction in the plane.

A function $f$ is called regular in a domain $D$ if it is differentiable at every point of $D$. A function regular on the whole complex plane is called entire.

Since $\Delta z\to0$ can proceed in any direction, the limit must agree along the real axis and along the imaginary axis. Write $f=u+iv$, where $u,v$ are the real and imaginary parts of $f$ as functions of $(x,y)$. For $\Delta z=\Delta x$ (along the real axis):

$$f'(z)=\lim_{\Delta x\to0}\frac{u(x+\Delta x,y)-u(x,y)}{\Delta x}+i\lim_{\Delta x\to0}\frac{v(x+\Delta x,y)-v(x,y)}{\Delta x}=u_x+iv_x.$$

For $\Delta z=i\Delta y$ (along the imaginary axis):

$$f'(z)=\frac1i\lim_{\Delta y\to0}\frac{u(x,y+\Delta y)-u(x,y)}{\Delta y}+\lim_{\Delta y\to0}\frac{v(x,y+\Delta y)-v(x,y)}{\Delta y}=v_y-iu_y.$$

Both limits are the same number $f'(z)$, so their real and imaginary parts agree:

$$u_x=v_y,\qquad u_y=-v_x.$$

These are the Cauchy-Riemann equations — a necessary condition for complex differentiability at a point. Unlike the real case, where differentiability at a single point is a fairly weak requirement, regularity (differentiability at every point of a domain at once) turns out to be a surprisingly rigid constraint — the subject of the next section, while the Cauchy-Riemann equations will be needed again in §3.

2. Taylor Series and Uniqueness

Unlike the real case, complex differentiability at every point of a domain implies the existence of derivatives of all orders — and representability of the function as a Taylor series in a neighborhood of any point of the domain of regularity:

$$f(z)=f(z_0)+f'(z_0)(z-z_0)+f''(z_0)\frac{(z-z_0)^2}{2}+f'''(z_0)\frac{(z-z_0)^3}{3!}+\dots$$

The proof of this fact is not quick and lies outside the scope of this material — but it is exactly what turns regularity from a local condition into an extremely rigid global one: knowing the function on an arbitrarily small neighborhood of a point lets you recover its values on the entire domain of regularity.

Uniqueness Theorem

Let $f$ and $g$ be regular in a domain $D$, and suppose there is a sequence $\{z_n\}\subset D$ converging to a point $w\in D$ on which $f(z_n)=g(z_n)$. Then $f=g$ throughout $D$.

Consequences: a regular function not identically zero has only finitely many zeros in any closed disk lying entirely within the domain of regularity; if two regular functions agree on some curve inside a shared domain of regularity, they agree throughout that domain. In particular, if two entire functions agree on a real interval $(-r;r)$, they agree on the whole plane $\C$.

This is used below for the complex exponential: a continuation of a real function to a regular function on $\C$, if it exists at all, is unique.

3. Cauchy's Integral Theorem and Formula

Let $f=u+iv$ be regular in a domain containing a closed contour $C$ and the region $D$ it bounds. Then

$$\oint_C f(z)\,dz=0.$$

Write $dz=dx+i\,dy$ and expand the product:

$$\oint_C f\,dz=\oint_C(u+iv)(dx+i\,dy)=\oint_C(u\,dx-v\,dy)+i\oint_C(v\,dx+u\,dy).$$

Green's formula applies to each of the two real contour integrals:

$$\oint_C P\,dx+Q\,dy=\iint_D\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dA.$$

For the real part ($P=u$, $Q=-v$):

$$\oint_C u\,dx-v\,dy=\iint_D(-v_x-u_y)\,dA.$$

For the imaginary part ($P=v$, $Q=u$):

$$\oint_C v\,dx+u\,dy=\iint_D(u_x-v_y)\,dA.$$

By the Cauchy-Riemann equations from §1 ($u_x=v_y$, $u_y=-v_x$), both integrands are identically zero: $-v_x-u_y=-v_x-(-v_x)=0$ and $u_x-v_y=0$. So both contour integrals vanish, and $\oint_C f\,dz=0$.

Geometrically, this means the integral $\int_{z_1}^{z_2}f\,dz$ along any path inside the domain of regularity depends only on the starting and ending point.

The theorem implies a far stronger fact — the Cauchy integral formula: the value of a regular function inside a contour is fully determined by its values on the contour itself.

Cauchy's Formula

Let $f$ be regular in a domain containing a closed contour $C$ and the region $D$ it bounds, and let $z_0\in D$. Then

$$f(z_0)=\frac1{2\pi i}\oint_C\frac{f(z)}{z-z_0}\,dz.$$

Proof sketch: the function $g(z)=f(z)/(z-z_0)$ is regular everywhere in $D$ except at the point $z_0$ itself. Cut out of $D$ a small disk of radius $\varepsilon$ centered at $z_0$: on the remaining ring between $C$ and the circle $C_\varepsilon$, $g$ is now regular everywhere. Green's formula also works on a domain with a "hole" (the boundary just needs to be oriented correctly — the outer contour counterclockwise, the inner one clockwise), so the theorem proved above gives $\oint_C g\,dz=\oint_{C_\varepsilon}g\,dz$ for any sufficiently small $\varepsilon$. Parametrizing $z=z_0+\varepsilon e^{i\theta}$:

$$\oint_{C_\varepsilon}\frac{f(z)}{z-z_0}\,dz=\int_0^{2\pi}\frac{f(z_0+\varepsilon e^{i\theta})}{\varepsilon e^{i\theta}}\,i\varepsilon e^{i\theta}\,d\theta=i\int_0^{2\pi}f(z_0+\varepsilon e^{i\theta})\,d\theta.$$

As $\varepsilon\to0$, the integrand tends to $f(z_0)$ uniformly in $\theta$ (regularity implies continuity), so the whole integral tends to $i\cdot2\pi f(z_0)$. The left-hand side $\oint_C g\,dz$ doesn't depend on $\varepsilon$ at all — so that is its value: $\oint_C\frac{f(z)}{z-z_0}\,dz=2\pi if(z_0)$, as claimed.

The formula reconstructs the function at any interior point from boundary data alone — and, differentiated with respect to $z_0$ under the integral sign, it is exactly what underlies the proof (omitted in §2) that a regular function is representable as a Taylor series.

4. The Complex Exponential and Euler's Formula

Define the complex exponential by the same power series as the real one:

$$\exp(z)=1+z+\frac{z^2}{2}+\frac{z^3}{3!}+\dots$$

The series converges at every point $z\in\C$: the partial sums form a Cauchy sequence, since for $n

The exponential can also be written as a limit:

$$e^z=\lim_{n\to\infty}\left(1+\frac zn\right)^n.$$

Take $z=i\varphi$, a purely imaginary number. On one hand, by definition $e^{i\varphi}=\lim_{n\to\infty}(1+i\varphi/n)^n$. On the other — the known limits $\cos(\varphi/n)=1-\varphi^2/2n^2+o(1/n^2)$ and $\sin(\varphi/n)=\varphi/n+o(1/n)$ give $\cos(\varphi/n)+i\sin(\varphi/n)=1+i\varphi/n+o(1/n)$ — that is, these two expressions differ by a quantity of order $o(1/n)$. A careful estimate (via the inequality $|z^n-w^n|\le n|z-w|\max\{|z|^{n-1},|w|^{n-1}\}$) shows that after raising to the $n$-th power this difference vanishes in the limit, and

$$e^{i\varphi}=\lim_{n\to\infty}\big(\cos(\varphi/n)+i\sin(\varphi/n)\big)^n=\cos\varphi+i\sin\varphi$$

— the last equality holds because multiplying $n$ equal numbers on the unit circle adds their arguments while their moduli stay equal to one. This is Euler's formula. At $\varphi=\pi$ it gives the famous identity

$$e^{i\pi}+1=0,$$

linking five fundamental constants in a single equation. Splitting the series for $e^{i\varphi}$ into real and imaginary parts gives the Taylor series for sine and cosine:

$$\cos\varphi=1-\frac{\varphi^2}{2!}+\frac{\varphi^4}{4!}-\dots,\qquad\sin\varphi=\varphi-\frac{\varphi^3}{3!}+\frac{\varphi^5}{5!}-\dots$$

The point on the circle is dragged by angle $\varphi$; the polyline is the partial sums $S_0,S_1,\dots,S_{10}$ of the series for $e^{i\varphi}$, spiraling toward the point $\cos\varphi+i\sin\varphi$ on the circle.

5. What's Next

Multiplying by a complex number is a rotational homothety (material «Complex Numbers: Three Constructions of One Field»). Regularity at a point where $f'(z_0)\ne0$ works much the same way: near $z_0$, the function $f$ is approximated by its linear part, $f(z)\approx f(z_0)+f'(z_0)(z-z_0)$ — the very same similarity, only with the coefficient $a=f'(z_0)$ now varying from point to point.

Similarities have a characteristic property — they preserve the angles between intersecting lines. What that implies for curves passing through $z_0$, and how to use it to construct explicit geometric maps of the plane, is examined in the material «Conformality and Fractional-Linear Transformations».