Categories: 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 neighbourhood 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 extremely “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}

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

\begin{aligned} \oint_C f(z) \:dz &= \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) \:dz &= - \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. Q.E.D.

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}

Proof. Thanks to the integral theorem, we know that the shape and size of $$C$$ is 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}

Q.E.D.

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}

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

\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 $$n$$. Q.E.D.

## Residue theorem

A function $$f(z)$$ is meromorphic if it is holomorphic except in a finite number of simple poles, which are points $$z_p$$ where $$f(z_p)$$ diverges, but where the product $$(z - z_p) f(z)$$ is non-zero and still holomorphic close to $$z_p$$.

The residue $$R_p$$ of a simple pole $$z_p$$ is defined as follows, and represents the rate at which $$f(z)$$ diverges close to $$z_p$$:

\begin{aligned} \boxed{ R_p = \lim_{z \to z_p} (z - z_p) f(z) } \end{aligned}

Cauchy’s residue theorem generalizes Cauchy’s integral theorem to meromorphic functions, and states that the integral of a contour $$C$$ depends on the simple poles $$p$$ it encloses:

\begin{aligned} \boxed{ \oint_C f(z) \dd{z} = i 2 \pi \sum_{p} R_p } \end{aligned}

Proof. From the definition of a meromorphic function, we know that we can decompose $$f(z)$$ like so, where $$h(z)$$ is holomorphic and $$p$$ are all its poles:

\begin{aligned} f(z) = h(z) + \sum_{p} \frac{R_p}{z - z_p} \end{aligned}

We integrate this over a contour $$C$$ which contains all poles, and apply both Cauchy’s integral theorem and Cauchy’s integral formula to get:

\begin{aligned} \oint_C f(z) \dd{z} &= \oint_C h(z) \dd{z} + \sum_{p} R_p \oint_C \frac{1}{z - z_p} \dd{z} = \sum_{p} R_p \: 2 \pi i \end{aligned}

Q.E.D.

This theorem might not seem very useful, but in fact, thanks to some clever mathematical magic, it allows us to evaluate many integrals along the real axis, most notably Fourier transforms. It can also be used to derive the Kramers-Kronig relations.