Theorem 8.1.1.
Suppose \(f(z) = \s c_k \left( z - z_0 \right)^k\) has radius of convergence \(R>0\text{.}\) Then \(f\) is holomorphic in \(D[z_0, R]\text{.}\)
Section 8.1 Power Series and Holomorphic Functions
Here is the first (and easier) half of the first goal we just announced.
Proof.
Corollary 7.4.9 says that \(f\) is continuous in \(D[z_0, R]\text{.}\) Given any closed piecewise smooth path \(\gamma \subset
D[z_0, R]\text{,}\) Corollary 7.4.10 gives \(\int_\gg f = 0\text{.}\) Now apply Morera’s theorem (Corollary 5.2.1).
A special case of this result concerns power series with infinite radius of convergence: those represent entire functions.
Now that we know that power series are differentiable in their regions of convergence, we can ask how to find their derivatives. The next result says that we can simply differentiate the series term by term.
Theorem 8.1.2.
Suppose \(f(z) = \s c_k ( z - z_0 )^k\) has radius of convergence \(R>0\text{.}\) Then
\begin{equation*}
f'(z) \ = \ \sum_{ k \geq 1 } k \, c_k ( z - z_0 )^{k-1}
\;\text{ for any } \;z\in D[z_0, R] \, \text{,}
\end{equation*}
and the radius of convergence of this power series is also \(R\text{.}\)
Proof.
If \(z\in D[z_0,R]\) then \(|z-z_0|\lt R\text{,}\) so we can choose \(R_1\) so that \(|z-z_0|\lt R_1\lt R\text{.}\) Then the circle \(\gg:=C[z_0,R_1]\) lies in \(D[z_0,R]\) and \(z\) is inside \(\gg\text{.}\) Since \(f\) is holomorphic in \(D[z_0,R]\) we can use Cauchy’s Integral Formula for \(f'\) (Theorem 5.1.1), as well as Corollary 7.4.10:
\begin{align*}
f'(z) \amp \ = \ \frac{1}{2 \pi i} \int_\gg
\frac{f(w)}{(w-z)^2} \, \diff{w} \ = \ \frac{1}{2 \pi i} \int_\gg
\frac{1}{(w-z)^2} \s c_k (w-z_0)^k \, \diff{w}\\
\amp \ = \ \s c_k \, \frac{1}{2 \pi i} \int_\gg
\frac{(w-z_0)^k}{(w-z)^2} \, \diff{w} \ = \ \s c_k
\left.\frac{d}{\diff{w}} (w-z_0)^k\right|_{w=z}\\
\amp \ = \ \sum_{ k \ge 1 } k \, c_k (z-z_0)^{k-1} \text{.}
\end{align*}
Note that we used Theorem 5.1.1 again in the penultimate step, but now applied to the function \((z-z_0)^k\text{.}\)
The last statement of the theorem is easy to show: the radius of convergence of \(f'(z)\) is at least \(R\) (since we have shown that the series for \(f'\) converges whenever \(\abs{z-z_0}\lt R\)), and it cannot be larger than \(R\) by comparison to the series for \(f(z)\text{,}\) since the coefficients for \((z-z_0) \, f'(z)\) are larger than the corresponding ones for \(f(z)\text{.}\)
Example 8.1.3.
In Example 7.4.8, we showed that \(f\) converges in \(\C\text{.}\) We claim that \(f(z) = \exp(z)\text{,}\) in analogy with the real exponential function. First, by Theorem 8.1.2,
\begin{equation*}
f'(z) \ = \ \frac{d}{\diff{z}} \sum_{k \geq 0} \frac{z^k}{k!}
\ = \ \sum_{k \geq 1} \frac{z^{k-1}}{(k-1)!} \ = \ \sum_{k \geq 0}
\frac{z^k}{k!} \ = \ f(z)\, \text{.}
\end{equation*}
Thus
\begin{align*}
\frac{ d }{ \diff{z} } \frac{ f(z) }{ \exp(z) }
\ \amp= \ \frac{ d }{ \diff{z} } \left( f(z) \exp(-z) \right)
\ = \ f'(z) \exp(-z) - f(z) \exp(-z)\\
\amp = \ 0 \, \text{,}
\end{align*}
and so, by Theorem 2.4.2, \(\frac{ f(z) }{ \exp(z) }\) is constant. Evaluating at \(z=0\) gives that this constant is 1, and so \(f(z) = \exp(z)\text{.}\)
Example 8.1.4.
We can use the power series expansion for \(\exp(z)\) to find power series for the trigonometric functions. For instance,
\begin{align*}
\sin z \amp \ = \ \frac{1}{2i}\left( \exp(iz) - \exp(-iz)
\right) \ = \ \frac{1}{2i}\left( \sum_{k\geq
0}\frac{(iz)^k}{k!} - \sum_{k\geq 0}\frac{(-iz)^k}{k!}
\right)\\
\amp \ = \ \frac{1}{2i} \sum_{k\geq 0}\frac{1}{k!}\left(
(iz)^k - (-1)^k(iz)^k \right) \ = \ \frac{1}{2i} \sum_{k\geq 0
\text{ odd } }\frac{2(iz)^k}{k!}\\
\amp \ = \ \frac{1}{i} \sum_{j\geq
0}\frac{(iz)^{2j+1}}{(2j+1)!} \ = \ \sum_{j\geq 0}\frac{i^{2j}
\, z^{2j+1}}{(2j+1)!} \ = \ \sum_{j\geq
0}\frac{(-1)^j}{(2j+1)!} \, z^{2j+1}\\
\amp \ = \ z - \frac{z^3}{3!} + \frac{z^5}{5!} -
\frac{z^7}{7!} + \cdots \,\text{.}
\end{align*}
Note that we are allowed to rearrange the terms of the two added sums because the corresponding series have infinite radii of convergence.
Naturally, Theorem 8.1.2 can be repeatedly applied to \(f'\text{,}\) then to \(f''\text{,}\) and so on. The various derivatives of a power series can also be seen as ingredients of the series itself—this is the statement of the following Taylor series expansion.
1
Named after Brook Taylor (1685–1731).
Corollary 8.1.5.
Suppose \(f(z) = \s c_k ( z - z_0 )^k\) has a positive radius of convergence. Then
\begin{equation*}
c_k \ = \ \frac{ f^{(k)} (z_0) }{ k! } \,\text{.}
\end{equation*}
Proof.
For starters, \(f(z_0)=c_0\text{.}\) Theorem 8.1.2 gives \(f'(z_0)=c_1\text{.}\) Applying the same theorem to \(f'\) gives
\begin{equation*}
f''(z) \ = \ \sum_{ k \geq 2 } k (k-1) \, c_k \left( z - z_0
\right)^{k-2}
\end{equation*}
and so \(f''(z_0) = 2 \, c_2\text{.}\) A quick induction game establishes \(f'''(z_0) = 6 \, c_3\text{,}\) \(f''''(z_0) = 24 \, c_4\text{,}\) etc.
Taylor’s formula shows that the coefficients of any power series converging to \(f\) on some open disk \(D\) can be determined from the function \(f\) restricted to \(D\text{.}\) It follows immediately that the coefficients of a power series are unique:
Corollary 8.1.6.
If \(\sum_{k\ge0}c_k \, (z-z_0)^k\) and \(\sum_{k\ge0}d_k \, (z-z_0)^k\) are two power series that both converge to the same function on an open disk centered at \(z_0\text{,}\) then \(c_k=d_k\) for all \(k \ge 0\text{.}\)
Example 8.1.7.
We’d like to compute a power series expansion for \(f(z) = \exp(z)\) centered at \(z_0 = \pi\text{.}\) Since
\begin{equation*}
f^{ (k) } (z_0) \ = \ \biggl. \exp(z) \biggr|_{ z=\pi } \ = \ e^\pi\text{,}
\end{equation*}
Corollary 8.1.5 suggests that this power series is
\begin{equation*}
\s \frac{ e^\pi }{ k! } \, (z-\pi)^k\text{,}
\end{equation*}
which converges for all \(z \in \C\) (essentially by Example 7.4.8).
We now turn to the second cornerstone result of this section, that a holomorphic function can be locally represented by a power series.
Theorem 8.1.8.
Suppose \(f\) is a function holomorphic in \(D[z_0, R]\text{.}\) Then \(f\) can be represented as a power series centered at \(z_0\text{,}\) with a radius of convergence \(\ge R\text{:}\)
\begin{equation*}
f(z) \ = \ \s c_k ( z - z_0 )^k \qquad \text{ with } \qquad
c_k \ = \ \frac 1 {2 \pi i} \int_\gg \frac{ f(w) }{
(w-z_0)^{k+1} } \, \diff{w} \, \text{,}
\end{equation*}
where \(\gg\) is any positively oriented, simple, closed, piecewise smooth path in \(D[z_0, R]\) for which \(z_0\) is inside \(\gg\text{.}\)
Proof.
Let \(g(z) := f(z+z_0)\text{;}\) so \(g\) is a function holomorphic in \(D[0,R]\text{.}\) Given \(z \in D[0,R]\text{,}\) let \(r := \frac{ |z|+R }{ 2 }\text{.}\) By Cauchy’s Integral Formula (Theorem 4.4.5),
\begin{equation*}
g (z) \ = \ \frac 1 {2 \pi i} \int_{C[0,r]} \frac{ g(w) }{ w-z }
\, \diff{w} \, \text{.}
\end{equation*}
The factor \(\frac 1 {w-z}\) in this integral can be expanded into a geometric series (note that \(w \in C[0,r]\) and so \(| \frac z w | \lt 1\)):
\begin{equation*}
\frac 1 {w-z} \ = \ \frac 1 w \, \frac 1 { 1 - \frac z w } \ = \
\frac 1 w \s \left( \frac z w \right)^k
\end{equation*}
which converges uniformly in the variable \(w \in C[0,r]\) by Exercise 7.5.30. Hence Proposition 7.3.6 applies:
\begin{align*}
g (z) \amp \ = \ \frac{1}{2 \pi i} \int_{C[0,r]} \frac{
g(w) }{ w-z } \, \diff{w}\\
\amp \ = \ \frac{1}{2 \pi i} \int_{C[0,r]}
g(w) \, \frac{1}{w} \s \left( \frac z w \right)^k \diff{w}\\
\amp \ = \
\s \left( \frac{1}{2 \pi i} \int_{C[0,r]} \frac{g(w)}{w^{k+1}}
\, \diff{w} \right) z^k \text{.}
\end{align*}
Now, since \(f(z) = g(z-z_0)\text{,}\) we apply a change of variables to obtain
\begin{equation*}
f(z) \ = \ \s \left( \frac 1 {2 \pi i} \int_{C[z_0, r]} \frac{
f(w) }{ (w-z_0)^{k+1} } \, \diff{w} \right) (z-z_0)^k \text{.}
\end{equation*}
The only differences of this right-hand side to the statement of the theorem are the paths we’re integrating over. However, by Cauchy’s Theorem 4.3.4,
\begin{equation*}
\int_{C[z_0, r]} \frac{ f(w) }{ (w-z_0)^{k+1} } \, \diff{w} \ = \
\int_\gg \frac{ f(w) }{ (w-z_0)^{k+1} } \, \diff{w} \, \text{.}
\end{equation*}
We note a remarkable feature of our proof: namely, if we are given a holomorphic function \(f: G \to \C\) and are interested in expanding \(f\) into a power series centered at \(z_0 \in G\text{,}\) then we may maximize the radius of convergence \(R\) of this power series, in the sense that its region of convergence reaches to the boundary of \(G\text{.}\) Let’s make this precise.
Definition 8.1.9.
For a region \(G \subseteq \C\) and a point \(z_0 \in G\text{,}\) we define the distance of \(z_0\) to \(\partial G\), the boundary of \(G\text{,}\) as the greatest lower bound of \(\{ |z-z_0| : \, z \in
\partial G \}\text{;}\) if this set is empty, we define the distance of \(z_0\) to \(\partial G\) to be \(\infty\text{.}\)
What we have proved above, on the side, is the following.
Corollary 8.1.10.
If \(f: G \to \C\) is holomorphic and \(z_0 \in G\text{,}\) then \(f\) can be expanded into a power series centered at \(z_0\) whose radius of convergence is at least the distance of \(z_0\) to \(\partial G\text{.}\)
Example 8.1.11.
Consider \(f: \C \setminus \{ \pm i \} \to \C\) given by \(f(z) := \frac 1 { z^2 + 1 }\) and \(z_0 = 0\text{.}\) Corollary 8.1.10 says that the power series expansion of \(f\) at 0 will have radius of convergence 1. (Actually, it says this radius is at least 1, but it cannot be larger since \(\pm i\) are singularities of \(f\text{.}\)) In fact, we can use a geometric series to compute this power series:
\begin{equation*}
f(z) \ = \ \frac 1 { z^2 + 1 } \ = \ \s \left( -z^2 \right)^k \ = \
\s (-1)^k \, z^{ 2k } \text{,}
\end{equation*}
with radius of convergence 1.
Corollary 8.1.10 is yet another example of a result that is plainly false when translated into \(\R\text{;}\) see Exercise 8.4.6.
Comparing the coefficients of the power series obtained in Theorem 8.1.8 with those in Corollary 8.1.5, we arrive at the long-promised extension of Theorem 4.4.5 and Theorem 5.1.1.
Corollary 8.1.12.
Suppose \(f\) is holomorphic in the region \(G\) and \(\gg\) is a positively oriented, simple, closed, piecewise smooth path, such that \(w\) is inside \(\gg\) and \(\gg \sim_G 0\text{.}\) Then
\begin{equation*}
f^{(k)} (w) \ = \ \frac{k!}{2 \pi i} \int_\gg \frac{ f(z) }{
(z-w)^{k+1} } \, \diff{z} \, \text{.}
\end{equation*}
Corollary 8.1.12 combined with our often-used Proposition 4.1.8(d) gives an inequality which is often called Cauchy’s Estimate:
Corollary 8.1.13.
Suppose \(f\) is holomorphic in \(D[w,R]\) and \(|f(z)| \leq M\) for all \(z \in D[w,R]\text{.}\) Then
\begin{equation*}
\left| f^{(k)} (w) \right| \ \leq \ \frac{ k! \, M }{ R^k } \,\text{.}
\end{equation*}
Proof.
\begin{align*}
\left| f^{(k)} (w) \right| \ \amp = \ \left| \frac{k!}{2
\pi i} \int_{ C[w,r] } \frac{ f(z) }{ (z-w)^{k+1} } \,
\diff{z} \right| \\\
\amp \leq \ \frac{k!}{2 \pi} \max_{z \in C[w,r]
} \left| \frac{ f(z) }{ (z-w)^{k+1} } \right| \length ( C[w,r]
)\\
\amp \leq \ \frac{k!}{2 \pi} \, \frac{ M }{ r^{k+1} } \,
2 \pi r\\
\amp \ = \ \frac{ k! \, M }{ r^k } \, \text{.}
\end{align*}
A key aspect of this section is worth emphasizing: namely, we have developed an alternative characterization of what it means for a function to be holomorphic. In Chapter 2, we defined a function to be holomorphic in a region \(G\) if it is differentiable at each point \(z_0 \in G\text{.}\) We now define what it means for a function to be analytic in \(G\text{.}\)
Definition 8.1.14.
Let \(f: G \to \C\) and \(z_0 \in G\text{.}\) If there exist \(R>0\) and \(c_0, c_1, c_2, \ldots \in \C\) such that the power series
\begin{equation*}
\s c_k (z-z_0)^k
\end{equation*}
converges in \(D[z_0, R]\) and agrees with \(f(z)\) in \(D[z_0, R]\text{,}\) then \(f\) is analytic at \(z_0\text{.}\) We call \(f\) analytic in \(G\) if \(f\) is analytic at each point in \(G\text{.}\)
What we have proved in this section can be summed up as follows:
Theorem 8.1.15.
For any region \(G\text{,}\) the class of all analytic functions in \(G\) coincides with the class of all holomorphic functions in \(G\text{.}\)
While the terms holomorphic and analytic do not always mean the same thing, in the study of complex analysis they do and are frequently used interchangeably.
