Holomorphic Function

In complex analysis, a holomorphic function is a complex differentiable function. The condition of complex differentiability is very strong, and leads to an especially elegant theory of calculus for these functions.

A function \(f: \mathbb{R} \to \mathbb{R}\) is said to be differentiable at \(x\in \mathbb{R}\) if \[\lim_{\substack{h\in \mathbb{R} \\ h\to 0}} \frac{f(x+h) - f(x)}{h}\] exists. If the limit exists, its value is called \(f'(x)\).

Analogously, if \(f\) is a function \(\mathbb{C} \to \mathbb{C}\) (that is, \(f\) takes complex numbers as arguments and returns complex numbers as results), then \(f\) is said to be complex differentiable at \(z\in \mathbb{C}\) if \[\lim_{\substack{h\in \mathbb{C} \\ h\to 0}} \frac{f(z+h) - f(z)}{h}\] exists. Again, if the limit exists, its value is called \(f'(z)\). If \(f\) is complex differentiable at every \(z\in U \subset \mathbb{C}\), then \(f\) is said to be holomorphic on \(U\).

Although both conditions are notationally the same, the complex differentiability condition is actually much stronger than the differentiability condition for functions \(\mathbb{R} \to \mathbb{R}\).

This is because as \(h\) approaches \(0\) in the complex limit, it must do so from all directions, of which there are infinitely many. (In the real limit, there are only two directions from which \(h\) may approach \(0\): the positive direction and the negative direction.) Thus, if \(f: \mathbb{C} \to \mathbb{C}\) is complex differentiable at \(z\in \mathbb{C}\), then \[\lim_{\substack{t\in \mathbb{R} \\ t\to 0}} \frac{f(z+th) - f(z)}{th} \] exists, for all \(h\in \mathbb{C}\), and equals \(f'(z)\).

One can view \(\mathbb{C}\) as \(\mathbb{R}^2\) by identifying the complex number \(x+yi\) with the pair \((x,y)\). From this perspective, one may write \(f(z) = f(x,y) = u(x,y) + i \cdot v(x,y)\), where \(u(x,y) = \text{Re}(f(z))\) and \(v(x,y) = \text{Im}(f(z))\). But then the complex differentiability condition requires \[f'(z) = \lim_{\substack{t\in \mathbb{R} \\ t\to 0}} \frac{f(z+t \cdot 1) - f(z)}{t\cdot 1} = \lim_{\substack{t\in \mathbb{R} \\ t\to 0}} \left(\frac{u(x+t,y) - u(x,y)}{t} + i \cdot \frac{v(x+t,y) - v(x,y)}{t} \right)\] \[ = \frac{\partial u}{\partial x}\Big\vert_{(x,y)} + i \frac{\partial v}{\partial x} \Big\vert_{(x,y)}\] and \[f'(z) = \lim_{\substack{t\in \mathbb{R} \\ t\to 0}} \frac{f(z+t \cdot i) - f(z)}{t\cdot i} = -i \cdot \lim_{\substack{t\in \mathbb{R} \\ t\to 0}} \left(\frac{u(x,y+t) - u(x,y)}{t} + i \cdot \frac{v(x,y+t) - v(x,y)}{t} \right)\] \[ = -i \cdot \left( \frac{\partial u}{\partial y} + i \cdot \frac{\partial v}{\partial y} \right) = \frac{\partial v}{\partial y}\Big\vert_{(x,y)} - i \cdot \frac{\partial u}{\partial y} \Big\vert_{(x,y)}\]

Setting these equal proves that wherever \(f\) is complex differentiable, it satisfies \[\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}\] \[\frac{\partial u}{\partial y} = - \frac{\partial v}{\partial x}\] These are the Cauchy-Riemann equations, a system of differential equations that every holomorphic function must satisfy on its domain. Hence, there are severe constraints on how a holomorphic function may behave; namely, it may only behave as these equations prescribe.

Perhaps the most important property of holomorphic functions is their analyticity. This means that, any holomorphic function has an expansion as a power series.

Suppose \(f: U \to \mathbb{C}\) is holomorphic, with \(U\) an open set. Then there exists a sequence of complex numbers \(\{a_n\}_{n\ge 0}\) such that

\[f(z) = \sum_{n=0}^{\infty} a_n (z-z_0)^n\]

within the radius of convergence about \(z_0\) (i.e., it converges for all \(z\) in some disk centered at \(z_0\)) and equals \(f(z)\) when it converges. Specifically, \(a_n = \frac{f^{(n)}(z_0)}{n!}\).

Let \(\gamma\) be a closed curve. Using the Cauchy integral formula and geometric series, we have

\[\begin{align} f(z) &= \frac{1}{2\pi i} \int_\gamma \frac{f(\zeta)}{\zeta-z} \ d\zeta \\ &= \frac{1}{2\pi i} \int_\gamma \frac{f(\zeta)/(\zeta-z_0)}{1-\frac{z-z_0}{\zeta-z_0}} \ d\zeta \\ &= \frac{1}{2\pi i} \int_\gamma \frac{f(\zeta)}{\zeta-z_0} \sum_{n=0}^\infty \left( \frac{z-z_0}{\zeta-z_0} \right)^n \ d\zeta \\ &= \sum_{n=0}^\infty \left( \frac{1}{2\pi i} \int_\gamma \frac{f(\zeta)}{(\zeta-z_0)^{n+1}} \ d\zeta \right) (z-z_0)^n \\ &= \sum_{n=0}^\infty \frac{f^{(n)}(z_0)}{n!}(z-z_0)^n. \end{align}\]