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3.2 Holomorphic Functions

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In this section we define what it means for a complex function $f$ defined on a domain $D$ to be holomorphic. First, we describe what it means for a function to be differentiable at a point and recover the usual rules for calculating derivatives.

Definition

Fix a domain $D \subset \mathbb{C}$. A function $f : D \to \mathbb{C}$ is said to be differentiable at a point $z \in D$ if the limit \[ \lim_{h \to 0} \dfrac{f(z+h) - f(z)}{h} \] exists. When it does exist we write $f'(z)$ for the limit.

Example

Let $f(z)=z^2$ be defined on $\CC$. For any $z \in \C$ we have \[ \frac{f(z+h)-f(z)}{h} = \frac{ z^2 + 2zh + h^2 - z^2}{h} = \frac{ 2zh + h^2}{h} = 2z+h \] so \[ \lim_{h \to 0} \frac{f(z+h) - f(z)}{h} = \lim_{h \to 0} 2z + h = 2z \] and $f'(z) = 2z$ for all $z \in \CC$.

Example

Define $f : \C \to \C$ by $f(z) = \overline{z}$. It is not differentiable! This is because \[\frac{\overline{z+h} - \overline{z \vphantom{h}}}{h} = \frac{\overline{h}}{h}\] does not have a limit as $h \to 0$. Indeed, if $h = t$ is real then the difference quotient takes the value 1, whereas if $h = it$ with $t$ real then the difference quotient is $-1$.

All of the standard rules from real analysis for computing derivatives continue to hold in the complex case.

Theorem (Differentiation Rules)

Let $f,g$ be defined on a domain $D$ and differentiable at $z \in D$. The following all hold.

  1. (Sum Rule) $(f + g)'(z) = f'(z) + g'(z)$
  2. (Scalar Rule) $(cf)'(z) = c f'(z)$ for every $c \in \C$
  3. (Product Rule) $(f \cdot g)'(z) = f'(z) g(z) + f(z) g'(z)$
  4. (Quotient Rule) $\left( \dfrac{f}{g} \right)'(z) = \dfrac{f'(z) g(z) - f(z) g'(z)}{(g(z))^2}$ if $g(z) \ne 0$.
Proof:

The proofs are identical to those for functions of a single real variable. $\square$

Theorem (Chain Rule)

Let $f$ be defined on a domain $D$ and suppose $f$ is differentiable at $z \in D$. Suppose also that $g$ is defined on a domain $E$ that contains $f(z)$ and $g$ is differentiable at $f(z)$.

  1. (Chain Rule) $(g \circ f)'(z) = g'(f(z)) \, f'(z)$
Proof:

The proof is identical to the real variable proof. $\square$

Lemma

If $f : D \to \C$ is differentiable at $z \in D$ then $f$ is continuous at $z$.

Proof:

To show that $f$ is continuous at $z$, we need to show that $\lim\limits_{h \to 0} f(z+h) = f(z)$, i.e. $\lim\limits_{h \to 0} f(z+h)-f(z) =0$. But \[ \lim_{h\to 0} f(z+h)-f(z) = \lim_{h\to 0} \frac{f(z+h)-f(z)}{h} \cdot h = f'(z) \cdot 0 = 0 \] as required. $\square$

The derivative of $f$ at $z$ is a complex number. What can we understand about the function from knowing its derivative at a certain point?

Theorem (Linearization)

Fix $f : D \to \C$ differentiable at a point $z \in D$. For every $\epsilon > 0$ there is $\delta > 0$ such that \[ \left| f(z+h) - \left( f(z) + f'(z) \, h \right) \right| < |h| \epsilon \] for all $h \in \C$ with $|h| < \delta$.

Proof:

From the limit statement in the definition of differentiable we have for every $\epsilon > 0$ the existence of some $\delta > 0$ such that \[ \left| \frac{f(z+h) - f(z)}{h} - f'(z) \right| < \epsilon \] whenever $|h| < \delta$. Multiplying through by $|h|$ gives the desired inequality. $\square$

This allows us to think of $f'(z)$ as follows: it encodes approximately how the function $f$ stretches and rotates near $z$.

Although the definition of differentiability of a function in complex analysis is, essentially, the same as the definition in real analysis, we lose many of the geometrical interpretations of the derivative. For example, one cannot easily interpret $f'(z)$ as the gradient or slope of $f$ at $z$. As another example, in real analysis one can normally interpret points $x$ for which $f'(x)=0$ as turning points or local maxima or minima of $f$. The notion of a local maximum or local minimum does not exist in complex analysis; this is because there is no natural ordering on the set of complex numbers.

Definition

Fix a domain $D \subset \mathbb{C}$. A function $f : D \to \mathbb{C}$ is called holomorphic on $D$ if it is differentiable at every point $z \in D$.

When $f : D \to \C$ is holomorhpic we can define a new function $f'$ on $D$ assigning to each point $z \in D$ the derivative $f'(z)$ there. This new function may itself happen to be holomorphic. If it is, we write its derivative as $f''(z)$ and so on.

Example

We saw in an earlier example that $f(z) = z^2$ has derivative $f'(z) = 2z$ for all $z \in \C$. Thus $f$ is a holomorphic function on $\C$. It is also a holomorphic function on any other domain $D \subset \C$.

Example

The function $f(z) = \dfrac{z}{z^2 - 1}$ is holomorphic on $\C \setminus \{i, -i\}$ and any domain $D$ contained in $\C \setminus \{i,-i\}$. It is not holomorphic, and not even defined on all of $\C \setminus \{i\}$ or $\C$.