Holomorphic functions §

Why complex differentiability is rigid §

A real derivative is a limit as $h \to 0$ along the real line — one direction. A complex derivative is a limit as $h \to 0$ from every direction in the plane. For the limit to exist, the difference quotient must approach the same value whether $h$ approaches $0$ along the real axis or along the imaginary axis.

Write $f = u + iv$. Along the real axis ($h = t \in \mathbb{R}$):

$\lim_{t \to 0} \frac{f(z_0 + t) - f(z_0)}{t} = u_x + i\, v_x.$

Along the imaginary axis ($h = it$):

$\lim_{t \to 0} \frac{f(z_0 + it) - f(z_0)}{it} = v_y - i\, u_y.$

Equating the two expressions (real part with real part, imaginary with imaginary) gives exactly the Cauchy–Riemann equations. So the equations are the statement that “the derivative does not depend on the direction of approach.”

Worked example §

Take $f(z) = z^2 = (x+iy)^2 = (x^2 - y^2) + i(2xy)$. Then $u = x^2 - y^2$ and $v = 2xy$.

$u_x = 2x, \quad v_y = 2x, \qquad u_y = -2y, \quad -v_x = -2y.$

Both Cauchy–Riemann equations hold everywhere, so $f$ is holomorphic on all of $\mathbb{C}$. Directly, $f'(z) = 2z$.

A function that fails §

Take $f(z) = \bar{z} = x - iy$, so $u = x$, $v = -y$. Then $u_x = 1$ but $v_y = -1$, so the first Cauchy–Riemann equation fails everywhere. Indeed $\bar{z}$ is nowhere complex-differentiable: approaching $0$ along the real axis gives quotient $1$, along the imaginary axis gives $-1$.

The key fact: holomorphic implies smooth §

On the real line, a function can be differentiable once and not twice (e.g. $f(x) = x^2 \sin(1/x^2)$ near $0$ with $f(0)=0$). In the complex plane this cannot happen: a function holomorphic on an open set is automatically $C^\infty$ and even real-analytic. The reason is that the Cauchy–Riemann equations make $u$ and $v$ harmonic, and harmonic functions are smooth.

Worked example: the exponential §

The power series $e^z = \sum_{n=0}^{\infty} z^n / n!$ has infinite radius of convergence, so it defines a holomorphic function on all of $\mathbb{C}$. Its derivative is term-by-term:

$\frac{d}{dz} \sum_{n=0}^{\infty} \frac{z^n}{n!} = \sum_{n=1}^{\infty} \frac{z^{n-1}}{(n-1)!} = \sum_{n=0}^{\infty} \frac{z^n}{n!} = e^z.$

So $(e^z)' = e^z$, exactly as in the real case.

Radius of convergence §

For a general power series $\sum a_n (z-z_0)^n$, the radius of convergence is $r = 1/\limsup |a_n|^{1/n}$ (with the usual conventions $0 \cdot \infty = 0$). Inside the disk the series converges absolutely and defines a holomorphic function; on the boundary the behavior can be anything.

Why the identity theorem holds §

Because $f$ equals a power series near any point, the zeros of $f$ are isolated unless $f \equiv 0$: if $f(z_0) = 0$ but $f$ is not identically zero, then $f(z) = (z - z_0)^m g(z)$ with $g(z_0) \neq 0$, so $f$ is nonzero in a punctured neighborhood of $z_0$. A set with an accumulation point therefore forces $f \equiv 0$. This is the “global behavior is controlled by values on any small open set” property stated in the overview.

Why the maximum modulus principle holds §

Suppose $|f|$ had a local maximum at an interior point $z_0$ with $f(z_0) \neq 0$. Then $g = 1/f$ is holomorphic near $z_0$ and $|g|$ has a local minimum there. But a non-constant holomorphic function cannot have a local extremum of modulus: expanding $f$ in a power series about $z_0$ shows that $|f|$ strictly increases in some direction and strictly decreases in another. Hence $f$ must be constant.

Consequence: minimum modulus and open mapping §

The maximum modulus principle applied to $1/f$ gives the minimum modulus principle. Together they imply the open mapping theorem: a non-constant holomorphic function maps open sets to open sets. This is the geometric heart of Conformal mapping.

Problem. Let $f$ be holomorphic on a connected open set $U$ and suppose $|f|$ is constant on $U$. Show that $f$ is constant.