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 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}

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}

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}

© Marcus R.A. Newman, a.k.a. "Prefetch". Available under CC BY-SA 4.0.