Categories: Complex analysis, Mathematics.

# Holomorphic function

In complex analysis, a complex function $f(z)$ of a complex variable $z$ is called holomorphic or analytic if it is complex differentiable in the vicinity of every point of its domain. This is a very strong condition.

As a result, holomorphic functions are infinitely differentiable and equal their Taylor expansion at every point. In physicists’ terms, they are very “well-behaved” throughout their domain.

More formally, a given function $f(z)$ is holomorphic in a certain region if the following limit exists for all $z$ in that region, and for all directions of $\Delta z$:

\begin{aligned} \boxed{ f'(z) = \lim_{\Delta z \to 0} \frac{f(z + \Delta z) - f(z)}{\Delta z} } \end{aligned}

We decompose $f$ into the real functions $u$ and $v$ of real variables $x$ and $y$:

\begin{aligned} f(z) = f(x + i y) = u(x, y) + i v(x, y) \end{aligned}

Since we are free to choose the direction of $\Delta z$, we choose $\Delta x$ and $\Delta y$:

\begin{aligned} f'(z) &= \lim_{\Delta x \to 0} \frac{f(z + \Delta x) - f(z)}{\Delta x} = \pdv{u}{x} + i \pdv{v}{x} \\ &= \lim_{\Delta y \to 0} \frac{f(z + i \Delta y) - f(z)}{i \Delta y} = \pdv{v}{y} - i \pdv{u}{y} \end{aligned}

For $f(z)$ to be holomorphic, these two results must be equivalent. Because $u$ and $v$ are real by definition, we thus arrive at the Cauchy-Riemann equations:

\begin{aligned} \boxed{ \pdv{u}{x} = \pdv{v}{y} \qquad \pdv{v}{x} = - \pdv{u}{y} } \end{aligned}

Therefore, a given function $f(z)$ is holomorphic if and only if its real and imaginary parts satisfy these equations. This gives an idea of how strict the criteria are to qualify as holomorphic.

## Integration formulas

Holomorphic functions satisfy Cauchy’s integral theorem, which states that the integral of $f(z)$ over any closed curve $C$ in the complex plane is zero, provided that $f(z)$ is holomorphic for all $z$ in the area enclosed by $C$:

\begin{aligned} \boxed{ \oint_C f(z) \dd{z} = 0 } \end{aligned}

Just like before, we decompose $f(z)$ into its real and imaginary parts:

\begin{aligned} \oint_C f(z) \dd{z} &= \oint_C (u + i v) \dd{(x + i y)} = \oint_C (u + i v) \:(\dd{x} + i \dd{y}) \\ &= \oint_C u \dd{x} - v \dd{y} + i \oint_C v \dd{x} + u \dd{y} \end{aligned}

Using Green’s theorem, we integrate over the area $A$ enclosed by $C$:

\begin{aligned} \oint_C f(z) \dd{z} &= - \iint_A \pdv{v}{x} + \pdv{u}{y} \dd{x} \dd{y} + i \iint_A \pdv{u}{x} - \pdv{v}{y} \dd{x} \dd{y} \end{aligned}

Since $f(z)$ is holomorphic, $u$ and $v$ satisfy the Cauchy-Riemann equations, such that the integrands disappear and the final result is zero.

An interesting consequence is Cauchy’s integral formula, which states that the value of $f(z)$ at an arbitrary point $z_0$ is determined by its values on an arbitrary contour $C$ around $z_0$:

\begin{aligned} \boxed{ f(z_0) = \frac{1}{2 \pi i} \oint_C \frac{f(z)}{z - z_0} \dd{z} } \end{aligned}

Thanks to the integral theorem, we know that the shape and size of $C$ are irrelevant. Therefore we choose it to be a circle with radius $r$, such that the integration variable becomes $z = z_0 + r e^{i \theta}$. Then we integrate by substitution:

\begin{aligned} \frac{1}{2 \pi i} \oint_C \frac{f(z)}{z - z_0} \dd{z} &= \frac{1}{2 \pi i} \int_0^{2 \pi} f(z) \frac{i r e^{i \theta}}{r e^{i \theta}} \dd{\theta} = \frac{1}{2 \pi} \int_0^{2 \pi} f(z_0 + r e^{i \theta}) \dd{\theta} \end{aligned}

We may choose an arbitrarily small radius $r$, such that the contour approaches $z_0$:

\begin{aligned} \lim_{r \to 0}\:\: \frac{1}{2 \pi} \int_0^{2 \pi} f(z_0 + r e^{i \theta}) \dd{\theta} &= \frac{f(z_0)}{2 \pi} \int_0^{2 \pi} \dd{\theta} = f(z_0) \end{aligned}

Similarly, Cauchy’s differentiation formula, or Cauchy’s integral formula for derivatives gives all derivatives of a holomorphic function as follows, and also guarantees their existence:

\begin{aligned} \boxed{ f^{(n)}(z_0) = \frac{n!}{2 \pi i} \oint_C \frac{f(z)}{(z - z_0)^{n + 1}} \dd{z} } \end{aligned}

By definition, the first derivative $f'(z)$ of a holomorphic function exists and is:

\begin{aligned} f'(z_0) = \lim_{z \to z_0} \frac{f(z) - f(z_0)}{z - z_0} \end{aligned}

We evaluate the numerator using Cauchy’s integral theorem as follows:

\begin{aligned} f'(z_0) &= \lim_{z \to z_0} \frac{1}{z - z_0} \bigg( \frac{1}{2 \pi i} \oint_C \frac{f(\zeta)}{\zeta - z} \dd{\zeta} - \frac{1}{2 \pi i} \oint_C \frac{f(\zeta)}{\zeta - z_0} \dd{\zeta} \bigg) \\ &= \frac{1}{2 \pi i} \lim_{z \to z_0} \frac{1}{z - z_0} \oint_C \frac{f(\zeta)}{\zeta - z} - \frac{f(\zeta)}{\zeta - z_0} \dd{\zeta} \\ &= \frac{1}{2 \pi i} \lim_{z \to z_0} \frac{1}{z - z_0} \oint_C \frac{f(\zeta) (z - z_0)}{(\zeta - z)(\zeta - z_0)} \dd{\zeta} \end{aligned}

This contour integral converges uniformly, so we may apply the limit on the inside:

\begin{aligned} f'(z_0) &= \frac{1}{2 \pi i} \oint_C \Big( \lim_{z \to z_0} \frac{f(\zeta)}{(\zeta - z)(\zeta - z_0)} \Big) \dd{\zeta} = \frac{1}{2 \pi i} \oint_C \frac{f(\zeta)}{(\zeta - z_0)^2} \dd{\zeta} \end{aligned}

Since the second-order derivative $f''(z)$ is simply the derivative of $f'(z)$, this proof works inductively for all higher orders $n$.