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Math 382-0: Kursobjekt

Aaron Greicius

Section 1.9 Complex differentiation

Definition 1.9.1. Interior point.

Given a subset \(D\subseteq \C\text{,}\) an element \(z_0\in D\) is an interior point of \(D\) if there is an \(r> 0\) such that \(B_r(z_0)\subseteq D\text{.}\) The interior of \(D\text{,}\) denoted \(D^\circ\) is defined as the set of all interior points of \(D\text{.}\)

Remark 1.9.2. Interior points.

The following observations follows easily from the relevant definitions.
  1. If \(z_0\) is an interior point of \(D\text{,}\) then \(z_0\) is a limit point of \(D\text{,}\)
  2. If \(U\) is an open subset of \(\C\text{,}\) then all points of \(U\) are interior points: i.e., \(U^\circ=U\text{.}\)

Definition 1.9.3. Complex differentiation.

Given a complex function \(f\) and an interior point \(z_0\) of the domain of \(f\text{,}\) we say \(f\) is differentiable at \(z_0\) if the limit
\begin{equation*} \lim_{z\rightarrow z_0}\frac{f(z)-f(z_{0})}{z-z_{0}}, \text{ equivalently }\lim_{h\rightarrow 0}\frac{f(z_{0}+h)-f(z_{0})}{h}, \end{equation*}
exists. When this is the case we call the value of this limit the derivative of \(f\) at \(z_{0}\text{,}\) denoted \(f'(z_{0})\text{.}\)
The function \(f\) is holomorphic (or analytic) on an open set \(U\) if it is differentiable at all points in \(U\text{.}\) When \(f\) is holomorphic on \(U\text{,}\) the function
\begin{align*} f'\colon U \amp \rightarrow \C\\ z \amp \mapsto f'(z) \end{align*}
is called the derivative of \(f\text{.}\) We will also denote \(f'\) as \(\frac{df}{dz}\text{,}\) and will write \(\frac{d}{dz}\) to denote the derivative operation.

Remark 1.9.4. Holomorphic versus analytic.

Technically speaking, a function analytic on an open set \(U\) if for all \(z_0\in U\) we can represent \(f(z)=\sum_{n=0}^\infty a_n(z-z_0)^n\) as a convergent power series of all \(z\) in an open ball around \(z_0\text{.}\) One of the headline theorems of this class is that \(f\) is differentiable on \(U\) if and only if it is analytic (in the sense just described). Consequently, in the literature the terms ‘differentiable’, ‘holomorphic’, and ‘analytic’ are used pretty much interchangeably. However, we will refrain from using ‘analytic’ until the theorem in question is proved.

Remark 1.9.5. Differentiable at a limit point.

In light of the limit definition of differentiability, it is of course possible to define this notion at any limit point of the domain of a function. As it turns out, however, an openness condition like the one we adopt is the more natural setting in terms of further important theorems (e.g., Theorem 1.10.2).

Example 1.9.6. Elementary examples.

Prove that the given \(f\) is differentiable on the given open set \(U\) and provide a formula for \(f'\text{.}\)
  1. Constant function.
    \(f(z)=w_0\text{,}\) \(w_0\in \C\text{,}\) \(U=\C\text{.}\)
  2. Identity function.
    \(f(z)=z\text{,}\) \(U=\C\text{.}\)
  3. Squaring function.
    \(f(z)=z^2\text{,}\) \(U=\C\text{.}\)
  4. Reciprocal function.
    \(f(z)=1/z\text{,}\) \(U=\C-\{0\}\text{.}\)
Solution.
It is easy to see using the limit definition that \(\frac{d}{dz}(w)=0\) and \(\frac{d}{dz}(z)=1\text{.}\) We illustrate the definition only for \(f(z)=z^2\) and \(f(z)=1/z\text{.}\)
Given \(f(z)=z^2\) and any \(z_0\in \C\text{,}\) we have
\begin{align*} \lim\limits_{z\to z_0}\frac{f(z)-f(z_0)}{z-z_0} \amp = \lim\limits_{z\to z_0}\frac{z^2-z_0^2}{z-z_0} \\ \amp =\lim\limits_{z\to z_0}\frac{(z-z_0)(z+z_0)}{z-z_0}\\ \amp =\lim\limits_{z\to z_0}z+z_0\\ \amp = 2z_0\text{.} \end{align*}
This shows that \(\frac{d}{dz}(z^2)=2z\) for all \(z\text{.}\)
Now let \(f(z)=1/z\text{.}\) We have
\begin{align*} \lim\limits_{z\to z_0}\frac{f(z)-f(z_0)}{z-z_0} \amp = \lim\limits_{z\to z_0}\frac{1/z-1/z_0}{z-z_0}\\ \amp = \lim\limits_{z\to z_0}\frac{(z_0-z)/(zz_0)}{z-z_0}\\ \amp = \lim\limits_{z\to z_0}\frac{z_0-z}{zz_0(z-z_0)}\\ \amp = \lim\limits_{z\to z_0}-\frac{1}{zz_0}\\ \amp = -\frac{1}{z_0^2}\text{.} \end{align*}
This shows that \(\frac{d}{dz}(1/z)=-1/z^2\) for all \(z\text{.}\)

Proof.

Note that by definition, \(z_0\) is assumed to be an interior point of the domain of \(f\text{,}\) which implies that \(z_0\) is a limit point of the domain. We are thus justified in using the limit definition of continuity. We have
\begin{align*} \lim\limits_{z\to z_0}f(z) \amp = \lim\limits_{z\to z_0}\left(\frac{f(z)-f(z_0)}{z-z_0}\cdot (z-z_0)+f(z_0)\right) \\ \amp = f'(z_0)(0)+f(z_0)=f(z_0)\text{.} \end{align*}
Thus \(f\) is continuous at \(z_0\text{.}\)

Proof.

These statements are proved in essentially the same way as the corresponding statements from elementary calculus. We omit them here.

Example 1.9.9. Polynomials.

Let \(f(z)=a_nz^n+a_{n-1}z^{n-1}+\cdots a_1z+a_0\text{.}\) Show that \(f\) is differentiable on \(\C\) and compute a formula for \(f'\text{.}\)
Solution.
First observe that for all nonnegative integers \(n\) the function \(g(z)=z^n\) is differentiable and satisfies \(g'(z)=nz^{n-1}\text{.}\) This is shown easily by induction: the base case \(n=0\) is just the constant function \(g(z)=1\text{,}\) and for the induction step, we have
\begin{align*} \frac{d}{dz}(z^n)\amp = \frac{d}{dz}(z^{n-1}z) \\ \amp =\frac{d}{dz}(z^{n-1})z+z^{n-1}\frac{d}{dz}z\\ \amp = (n-1)z^{n-2}\, z+z^{n-1}\cdot 1\\ \amp = nz^{n-1}\text{.} \end{align*}

Proof.

This is left as a homework exercise.
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