3.2 Holomorphic Functions

Home | Assessment | Notes | Index | Worksheets | Blackboard

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 $$\underset{h\to 0}{lim}{\displaystyle \frac{f(z+h)-f(z)}{h}}$$ exists. When it does exist we write $f^{\prime }(z)$ for the limit.

Example

Let $f(z)=z^2$ be defined on $\mathbb{C}$. For any $z\in \mathbb{C}$ 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 $$\underset{h\to 0}{lim}\frac{f(z+h)-f(z)}{h}=\underset{h\to 0}{lim}2z+h=2z$$ and $f^{\prime }(z)=2z$ for all $z\in \mathbb{C}$.

Example

Define $f:\mathbb{C}\to \mathbb{C}$ by $f(z)=\bar{z}$. It is not differentiable! This is because $$\frac{\overline{z+h}-\overline{z\phantom{h}}}{h}=\frac{\bar{h}}{h}$$ 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) $(f+g{)}^{\prime }(z)=f^{\prime }(z)+g^{\prime }(z)$
2. (Scalar Rule) $(cf{)}^{\prime }(z)=c{f}^{\prime }(z)$ for every $c\in \mathbb{C}$
3. (Product Rule) $(f\cdot g{)}^{\prime }(z)=f^{\prime }(z)g(z)+f(z)g^{\prime }(z)$
4. (Quotient Rule) ${\left({\displaystyle \frac{f}{g}}\right)}^{\prime }(z)={\displaystyle \frac{f^{\prime }(z)g(z)-f(z)g^{\prime }(z)}{(g(z){)}^2}}$ if $g(z)\ne 0$.

Proof:

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

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 $f(z)$.

5. (Chain Rule) $(g\circ f{)}^{\prime }(z)=g^{\prime }(f(z))f^{\prime }(z)$

Proof:

The proof is identical to the real variable proof. $◻$

Lemma

If $f:D\to \mathbb{C}$ 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 $\underset{h\to 0}{lim}f(z+h)=f(z)$, i.e. $\underset{h\to 0}{lim}f(z+h)-f(z)=0$. But $$\underset{h\to 0}{lim}f(z+h)-f(z)=\underset{h\to 0}{lim}\frac{f(z+h)-f(z)}{h}\cdot h=f^{\prime }(z)\cdot 0=0$$ as required. $◻$

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 $f:D\to \mathbb{C}$ differentiable at a point $z\in D$. For every $ϵ>0$ there is $\delta >0$ such that $$|f(z+h)-(f(z)+f^{\prime }(z)h)|<|h|ϵ$$ for all $h\in \mathbb{C}$ with $|h|<\delta$.

Proof:

From the limit statement in the definition of differentiable we have for every $ϵ>0$ the existence of some $\delta >0$ such that $$|\frac{f(z+h)-f(z)}{h}-f^{\prime }(z)|<ϵ$$ whenever $|h|<\delta$. Multiplying through by $|h|$ gives the desired inequality. $◻$

This allows us to think of $f^{\prime }(z)$ 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 $f^{\prime }(z)$ as the gradient or slope of $f$ at $z$. As another example, in real analysis one can normally interpret points $x$ for which $f^{\prime }(x)=0$ 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 $f:D\to \mathbb{C}$ is holomorhpic we can define a new function $f^{\prime }$ on $D$ assigning to each point $z\in D$ the derivative $f^{\prime }(z)$ 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 $f^{\prime }(z)=2z$ for all $z\in \mathbb{C}$. Thus $f$ is a holomorphic function on $\mathbb{C}$. It is also a holomorphic function on any other domain $D\subset \mathbb{C}$.

Example

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