Section 1.17 Complex power series
Definition 1.17.1. Complex power series.
Given \(z_0\in \C\text{,}\) a (complex) power series centered at \(z_0\) in the variable (or unknown) \(z\) is an expression of the form
\begin{equation}
f(z)=\sum_{n=0}^\infty a_n(z-z_0)^n=a_0+a_1(z-z_1)+a_2(z-z_0)^2+\cdots\text{,}\tag{1.63}
\end{equation}
where \(a_n\in \C\text{.}\) We call \(z_0\) the center of the power series, and \(a_n\) the \(n\)-th coefficient of the series.
For any integer \(k\geq 0\text{,}\) the \(k\)-th partial sum of the power series \(f\) is the function \(f_k(z)\) defined as
\begin{equation*}
f_k(z)=\sum_{n=0}^ka_nz^n\text{.}
\end{equation*}
Given \(w\in \C\text{,}\) we say the power series \(f(z)=\sum_{n=0}^{\infty}a_n(z-z_0)^n\) converges (resp. diverges) at \(w\) if the infinite series \(f(w)=\sum_{n=0}^{\infty}a_n(w-z_0)^n\) converges (resp. diverges).
Recall the geometric series
\(f(z)=\sum_{n=0}^\infty z^n\) treated in
Example 1.5.11. We saw there that the series converges for all
\(z\) with
\(\abs{z}< 1\) and showed that for all such
\(z\) we have
\begin{equation}
f(z)=\sum_{n=0}^\infty z^n=\frac{1}{1-z}\text{.}\tag{1.64}
\end{equation}
We also showed that the series diverges for all \(z\) with \(\abs{z}> 0\) (as well as for all \(z\) with \(\abs{z}=1\)). Make this example your friend. It is both one of the simplest power series and one of the most fundamental, and will show up with an uncanny persistence.
The formula for the function
\(f(z)=\frac{1}{1-z}\) given in
(1.64) is an example of a
power series representation of
\(f\) on the open set
\(B_1(0)\text{.}\)
Definition 1.17.2. Power series representation.
Given a complex function \(f\) defined on the open set \(U\text{,}\) a power series representation (or expansion) of \(f\) on \(U\) is a power series \(\sum_{n=0}^\infty a_n(z-z_0)^n\) that satisfies
\begin{equation*}
f(z)=\sum_{n=0}^{\infty}a_n(z-z_0)^n
\end{equation*}
for all \(z\in U\text{.}\)
We will be able to generate many power series representations of functions from the geometric series, as the next example indicates.
Example 1.17.3. Power series representation.
Find a power series representation of the given function, centered at the given \(z_0\text{.}\) Let \(f(z)=\frac{z^3}{9-z^2}\text{.}\) Find power series representation of \(f\) centered at \(z_0=0\) on the open set \(U=B_3(0)\text{.}\)
Solution.
We have
\begin{align*}
f(z) \amp =\frac{z^3}{9}\cdot \frac{1}{1-(z/3)^2}\text{.}
\end{align*}
Since formula
(1.64) is valid for
all complex in
\(B_1(0)\text{,}\) we have
\begin{equation*}
\frac{1}{1-(z/3)^2}=\sum_{n=0}^\infty((z/3)^2)^n=\sum_{n=0}^\infty \frac{1}{3^{2n}}z^{2n}
\end{equation*}
for all \(z\) with \(\abs{(z/3)^2}< 1\text{,}\) and thus for all \(z\) with \(\abs{z}< 3\text{.}\) Using the product law for limits, it follows that
\begin{equation*}
f(z)=\frac{z}{9}\sum_{n=0}^\infty \frac{1}{3^{2n}}z^{2n}=\sum_{n=0}^\infty\frac{1}{3^{2n+2}}z^{2n+1}\text{.}
\end{equation*}
Theorem 1.17.5 details not only where a power series converges, but also
how it converges. The latter discussion touches on the notion of uniform convergence of sequences of functions, which we now officially introduce.
Definition 1.17.4. Sequences of functions.
Let \((f_n)_{n=0}^\infty\) be a sequence of complex functions of the form \(f_n\colon D\rightarrow \C\text{.}\)
-
Pointwise convergence.
The sequence \((f_n)\) is said to converge pointwise to a function \(f\colon D\rightarrow \C\) if for all \(z\in D\) we have \(f_n(z)\to f(z)\text{:}\) i.e., for all \(z\in D\) and \(\epsilon > 0\text{,}\) there exists an integer \(N\geq 0\) such that \(\abs{f_n(z)-f(z)}< \epsilon\) for all \(n\geq N\text{.}\) We write \(\lim\limits_{n\to \infty}f_n=f\) or \(f_n\to f\) when this is the case.
-
Uniform convergence.
The sequence \((f_n)\) is said to converge uniformly to a function \(f\colon D\rightarrow \C\) if for all \(\epsilon > 0\text{,}\) there exists an integer \(N\geq 0\) such that for all \(z\in D\) and for all \(n\geq N\text{,}\) \(\abs{f_n(z)-f(z)}< \epsilon\text{.}\)
Theorem 1.17.5. Power series convergence.
Let \(f(z)=\sum_{n=0}^\infty a_n (z-z_0)^n\) be a complex power series and suppose \(f\) converges at \(w_0\in \C\text{.}\) Let \(R=\abs{w_0-z_0}\text{.}\)
-
Absolute convergence.
For all \(z\in B_{R}(z_0)\) the infinite series \(f(z)\) converges absolutely.
-
Uniform convergence.
For all integers \(k\geq 0\text{,}\) let \(f_k(z)=\sum_{n=0}^ka_nz^n\) be the \(k\)-th partial sum of \(f\text{.}\) Given any \(r\) with \(0< r< R\) the sequence of partial sums \((f_k)_{k=0}^{\infty}\) converges to \(f\) uniformly on \(\overline{B}_r(z_0)\text{.}\)
Proof.
We first observe that since \(f(w_0)=\sum_{n=0}^\infty a_n(w_0-z_0)^n\) converges, we have
\begin{equation*}
\abs{a_n (w_0-z_0)^n}\to 0\text{,}
\end{equation*}
and this in turn implies that there is an \(M\geq 0\) such that \(\abs{a_n (w_0-z_0)^n}\leq M\) for all \(n\text{.}\)
Let \(w\in B_R(z_0)\text{.}\) We will show that
\begin{equation*}
\sum_{n=0}^\infty \abs{a_n(w-z_0)^n}
\end{equation*}
converges by comparing it to a certain geometric series. Namely, we have
\begin{align*}
\abs{a_n(w-z_0)^n} \amp =\abs{a_n(w_0-z_0)^n\left(\frac{w-z_0}{w_0-z_0}\right)^n}\\
\amp = \abs{a_n(w_0-z_0)^n}\abs{\frac{w-z_0}{R}}^n \\
\amp \leq M\abs{\frac{w-z_0}{R}}^n\text{,}
\end{align*}
where \(M\) is the bound on \(\abs{a_n(w_0-z_0)^n\) discussed above. Since \(w\in B_R(z_0)\text{,}\) we have \(\abs{(w-z_0)/r}< 1\text{,}\) and thus the geometric series
\begin{equation*}
\sum_{n=0}^\infty M \abs{\frac{w-z_0}{R}}^n
\end{equation*}
converges. It follows by the comparison test that \(\sum_{n=0}^\infty \abs{a_n(w-w_0)^n}\) converges. Thus \(f(w)\) is absolutely convergent.
First observe that for all \(z\in B_R(z_0)\) we have
\begin{align*}
\abs{f_k(z)-f(z)} \amp = \abs{f(z)-f_k(z)}\\
\amp = \abs{\sum_{n=0}^\infty a_n(z-w_0)^n-\sum_{n=k}^\infty a_n(z-w_0)^n}\\
\amp = \abs{\sum_{n=k+1}^\infty a_n(z-w_0)^n}
\end{align*}
for all integers \(k\geq 0\text{.}\) Fix any \(r\) with \(0 < r < R\text{.}\) Given any \(z\in B_r(z_0)\text{,}\) arguing exactly as above, we have
\begin{align*}
\abs{\sum_{n=k+1}^\infty a_n(z-w_0)^n}\amp \leq \sum_{n=k+1}^\infty \abs{a_n(z-w_0)^n}\\
\amp \leq \sum_{n=k+1}^\infty M\abs{\frac{z-z_0}{w_0-z_0}}^n\\
\amp = M\sum_{n=k+1}^\infty (r/R)^n\\
\amp = M(r/R)^k\sum_{n=0}^\infty (r/R)^n\\
\amp = M(r/R)^k\cdot \frac{1}{1-(r/R)}\\
\amp =(r/R)^k\cdot \frac{MR}{R-r}\text{,}
\end{align*}
since \(r/R< 1\text{.}\) Since \(M\text{,}\) \(R\text{,}\) and \(r\) are fixed constants, we see that
\begin{equation*}
(r/R)^k\cdot \frac{MR}{R-r}\to 0
\end{equation*}
as \(k\to \infty\text{.}\) It follows that for any \(\epsilon > 0\) we can find an integer \(N\geq 0\) such that \(k\geq N\) implies
\begin{equation*}
(r/R)^k\cdot \frac{MR}{R-r}< \epsilon\text{.}
\end{equation*}
We conclude that for all \(z\in B_{r}(z_0)\) we have
\begin{equation*}
\abs{f_k(z)-f(z)}\leq (r/R)^k\cdot \frac{MR}{R-r}< \epsilon
\end{equation*}
for all \(k\geq N\text{.}\) This shows that \(f_k\to f\) uniformly on \(B_r(z_0)\text{,}\) as claimed.
Corollary 1.17.6. Radius of convergence.
Let \(f(z)=\sum_{n=0}^\infty a_n (z-z_0)^n\) be a complex power series. There is a unique extended real number \(R\in [0,\infty)\cup \{\infty\}\) satisfying the properties below.
\(f\) converges for all \(z\) satisfying \(\abs{z-z_0}< R\text{.}\)
\(f\) diverges for all \(z\) satisfying \(\abs{z-z_0}> R\text{.}\)
Definition 1.17.7. Radius of convergence.
The
radius of convergence of a power series
\(f(z)=\sum_{n=0}^\infty a_nz^n\) is the unique extended real number
\(R\in [0,\infty)\cup \{\infty\}\) satisfying properties (i)-(ii) in
Corollary 1.17.6. The
circle of convergence of the power series
\(f\) is the set
\(\{z\in \C\colon \abs{z-z_0}=R\}\text{.}\)
Example 1.17.8. Expansions and radii of convergence.
Find a power series expansion of the given \(f\) centered at the given \(z_0\) and compute the radius of convergence of the expansion.
\(f(z)=1/z\text{,}\) \(z_0=i\)
\(f(z)=1/z\text{,}\) \(z_0=2\)
\(f(z)=1/(z^2+2)\text{,}\) \(z_0=0\)
Solution.
We can write
\begin{equation*}
\frac{1}{z}=\frac{1}{i+z-i}=\frac{1}{i}\cdot\frac{1}{1-i(z-i)}\text{.}
\end{equation*}
Since
\begin{equation*}
\frac{1}{1-w}=\sum_{n=0}^\infty w^n
\end{equation*}
for all \(w\in B_{1}(0)\text{,}\) we see that
\begin{equation}
\frac{1}{z}=\frac{1}{i}\sum_{n=0}^\infty(i(z-i))^n=\sum_{n=0}^\infty i^{n-1}(z-i)^n\tag{1.65}
\end{equation}
for all \(z\) satisfying \(\abs{i(z-i)}=\abs{z-i}< 1\text{.}\) Furthermore, since the geometric series diverges for all \(\abs{w}> 1\text{,}\) we see that the series \(\sum_{n=0}^\infty i^{n-1}(z-i)\) diverges for all \(z\) with \(\abs{z-i}> 1\text{.}\) We conclude that the radius of convergence is \(1\text{.}\)
Proceeding similarly as in (1), we have
\begin{equation*}
\frac{1}{z}=\frac{1}{2+(z-2)}=\frac{1}{2}\cdot\frac{1}{1+(z-2)/2}\text{,}
\end{equation*}
and thus, using the expansion \(\frac{1}{1+w}=\frac{1}{1-(-w)}=1-w+w^2+\cdots \text{,}\) we have
\begin{equation*}
\frac{1}{z}=\frac{1}{2}\sum_{n=0}^{\infty}(-1)^n(z-2)^n/2^{n}=\sum_{n=0}^\infty \frac{(-1)^n}{2^{n+1}}(z-2)^n\text{.}
\end{equation*}
Furthermore, reasoning just as in (1), we see that the radius of convergence of this power series is \(2\text{:}\) i.e., the series converges for all \(z\in B_2(2)\) and diverges for all \(z\) with \(\abs{z-2}> 2\text{.}\)
Since
\begin{equation*}
\frac{1}{z^2+4}=\frac{1}{4}\cdot\frac{1}{1+(z^2/4)}\text{,}
\end{equation*}
reasoning as above we see that
\begin{equation*}
\frac{1}{z^2+4}=\frac{1}{4}\sum_{n=0}^\infty (-1)^n(z^2/4)^n=\sum_{n=0}^\infty \frac{(-1)^n}{4^{n+1}}z^{2n}\text{,}
\end{equation*}
for all \(z\) satisfying \(\abs{z^2}/4 < 1\text{,}\) or equivalently \(\abs{z}< 2\text{.}\) Similarly since the series is seen to diverge if \(\abs{z}> 2\text{,}\) we conclude that the radius of convergence is \(2\text{.}\)
An important consequence of the uniform convergence of the partial sum polynonmial functions \(f_k(z)=\sum_{n=0}^ka_n(z-z_0)^n\) to \(f(z)=\sum_{n=0}^\infty a_n(z-z_0)^n\) is that the function \(f\) itself can be treated much like a polynomial: in particular, just as with polynomials we can differentiate and integrate \(f\) term by term.
Theorem 1.17.9. Power series integration and differentiation.
Let \(f(z)=\sum_{n=0}^\infty a_nz^n\) be a power series with radius of convergence \(R\text{.}\)
\(f\) is continuous on \(B_R(z_0)\text{.}\)
For any path \(\gamma\) lying in \(B_{R}(z_0)\text{,}\) we have
\begin{equation}
\int_\gamma f\, dz=\sum_{n=0}^\infty \int_\gamma a_n(z-z_0)\, dz\text{.}\tag{1.66}
\end{equation}
\(f\) is holomorphic on \(B_R(z_0)\) and
\begin{equation}
f'(z)=a_1+2a_2z+3a_3z^2+\cdots=\sum_{n=0}^\infty (n+1)a_{n+1}z^n\text{.}\tag{1.67}
\end{equation}
In other words, we a power series representation of \(f'\) by differentiating \(f\) term by term.
Proof.
We show directly that \(f\) is continuous at all \(z\in B_R(z_0)\text{.}\) Given \(z\in B_{R}(z_0)\) and \(\epsilon > 0\text{,}\) first pick an \(r> 0\) such that \(\abs{z-z_0}< r < R\text{.}\) Since the partial sum functions \(f_k(z)\) converge uniformly to \(f(z)\) on \(\overline{B}_r(z_0)\text{,}\) we can find an integer \(N\) such that \(\abs{f_k{w}-f(w)}< \epsilon/3\) for all \(w\in B_r(z_0)\) and all \(k\geq N\text{.}\) Furthermore, since \(f_N(z)\) is continuous (being a polynomial), we can find a \(delta\) such that \(\abs{w-z}< \delta\) implies \(\abs{f_N(w)-f_N(z)}< \epsilon/3\text{.}\) Shrinking \(\delta\) as necessary, we can further assume that \(\abs{w-z}< \delta\) implies \(w\in B_r(z_0)\text{.}\) We then conclude for all such \(w\) we have
\begin{align*}
\abs{f(w)-f(z)} \amp = \abs{f(w)-f_k(w)+(f_k(w)-f_k(z))+(f_k(z)-f(w))} \\
\amp \leq \abs{f(w)-f_k(w)}+\abs{f_k(w)-f_k(z)}+\abs{f_k(z)-f_k(z)}\\
\amp < \frac{\epsilon}{3}+\frac{\epsilon}{3}+\frac{\epsilon}{3}\\
\amp = \epsilon\text{.}
\end{align*}
This shows \(f\) is continuous at all \(z\in B_R(z_0)\text{.}\)
-
Let \(\gamma\colon [a,b]\rightarrow B_R(z)\) be a closed simple path. We show something more general: namely, if \((f_k)\) is any sequence of functions that converges uniformly to \(f\) on \(\gamma\text{,}\) then
\begin{equation*}
\int_\gamma f\, dz=\lim_{k\to \infty}\int_\gamma f_k\, dz\text{.}
\end{equation*}
The result of the theorem then follows by picking \(r< R\) such that \(\gamma\) lies within \(\overline{B}_r(z_0)\text{,}\) and taking \((f_k)\) to be the sequence of partial sum functions.
We now prove our more general claim. Given \(\epsilon > 0\text{,}\) pick \(N\) such that \(\abs{f_k(z)-f(z)}< \epsilon/L(\gamma)\) for all \(k\geq N\) , where \(L=\int_a^b\abs{\gamma'(t)}\, dt\) is the arc length of \(\gamma\text{.}\) It follows that for all \(k\geq N\) we have
\begin{align*}
\abs{\int_\gamma f\, dz-\int_\gamma f_k \, dz} \amp = \abs{\int_\gamma f(z)-f_k(z) \, dz} \\
\amp \leq \int_\gamma\abs{f(z)-f_k(z)}\, dz\\
\amp \leq \frac{\epsilon}{L(\gamma)}\cdot L(\gamma) \\
\amp = \epsilon\text{.}
\end{align*}
This proves
\begin{equation*}
\lim\limits_{k\to \infty}\int_\gamma f_k\, dz=\int_\gamma f\, dz\text{,}
\end{equation*}
as claimed.
For each integer
\(k\geq 0\text{,}\) let
\(f_k(z)=\sum_{n=0}^ka_n(z-z_0)^k\) be the
\(k\)-th partial sum of the power series. Interestingly, we will prove the claim by using our general result in (2) about limits of integrals, and the
\(n=1\) case of the
generalized Cauchy integral formula. In more detail, given any
\(w\in B_R(z_0)\text{,}\) pick an
\(r\) such that
\(w\in \overline{B}_r(z_0)\subseteq B_R(z_0)\) and pick an
\(s> 0\) such that
\(\gamma(t)=w+se^{it}\subseteq \overline{B}_r(z_0)\text{.}\) We then have
\begin{align*}
f'(w)\amp=\frac{1}{2\pi i}\int_\gamma \frac{f(z)}{z-w}\, dz \\
f_k'(w) \amp = \frac{1}{2\pi i}\int_\gamma \frac{f_k(z)}{z-w}\, dz \text{.}
\end{align*}
Since \(f_k(z)\to f(z)\) uniformly for all \(z\in \overline{B}_r(z_0)\text{,}\) it follows that \(\tfrac{f_k(z)}{2\pi i(z-w)}\to \tfrac{f(z)}{2\pi i(z-w)}\) uniformly on \(\gamma\text{.}\) Indeed, given any \(\epsilon > 0\text{,}\) pick \(N\) such that \(\abs{f_k(z)-f(z)}< 2\pi \epsilon s\) for all \(k\geq N\text{;}\) then for all \(z=w+se^{it}\) on \(\gamma\text{,}\) we have
\begin{align*}
\abs{\frac{f(z)}{2\pi i(z-w)}-\frac{f_k(z)}{2\pi i(z-w)}}\amp = \frac{\abs{f(z)-f_k(z)}}{2\pi\abs{z-w}} \\
\amp < \frac{2\pi \epsilon s}{2\pi s}\\
\amp < \epsilon\text{.}
\end{align*}
Lastly, since \(\tfrac{f_k(z)}{2\pi i(z-w)}\to \tfrac{f(z)}{2\pi i(z-w)}\) uniformly on \(\gamma\text{,}\) by the general result proved in (2) above , we have
\begin{align*}
f'(z) \amp =\int_{\gamma}\frac{f(z)}{2\pi i(z-w)}\, dz\\
\amp = \lim\limits_{k\to \infty} \int_{\gamma}\frac{f_k(z)}{2\pi i(z-w)}\, dz \\
\amp = \lim\limits_{k\to \infty}f_k'(z) \\
\amp = \lim\limits_{k\to \infty}\sum_{n=0}^kna_n(z-z_0)^{n-1} \amp (f_k(z)=\sum_{n=0}^k a_n(z-z_0)^n) \\
\amp = \sum_{n=0}^\infty na_n(z-z_0)^{n-1}\\
\amp = \sum_{n=0}^\infty (n+1)a_{n+1}(z-z_0)^{n}\text{,}
\end{align*}
as claimed.
Corollary 1.17.10. Radius of convergence.
Let \(f(z)=\sum_{n=0}^\infty a_n(z-z_0)^n\) be a power series centered at \(z_0\) with radius of convergence \(R\text{,}\) and let \(C\in \C\) be any fixed complex constant. The two power series
\begin{align*}
F(z) \amp =C+a_0z+\frac{a_1}{2}z^2+\cdots=C+\sum_{n=1}^\infty \frac{a_{n-1}}{n}(z-z_0)^n\\
g(z) \amp = a_1+2a_2(z-z_0)+3a_3(z-z_0)^3+\cdots =\sum_{n=0}^\infty (n+1)a_{n+1}(z-z_0)^n
\end{align*}
each have radius of convergence \(R\text{.}\) In other words, if the power series \(f(z)\) has radius of convergence \(R\text{,}\) then so does any formal antiderivative \(F(z)\) of \(f(z)\) and the formal derivative \(g(z)\) of \(f(z)\text{.}\)
Proof.
It is enough to show that taking the formal antiderivative of any power series does not change the radius of convergence, since \(F\) is a formal antiderivative of \(f\) and \(f\) is a formal antiderivative of \(g\text{.}\)
Assume the power series
\(F(z)\) has radius of convergence
\(R'\text{.}\) Pick any
\(w\in B_R(z_0)\text{,}\) and let
\(\gamma_{z_0,w}\) be the straight line path from
\(z_0\) to
\(z\text{,}\) which lies within
\(B_R(z_0)\text{.}\) By (2) of
Theorem 1.17.9, we have
\begin{align*}
\int_{\gamma_{z_0,w}} f(z)\, dz\amp =\sum_{n=0}^\infty \int_{\gamma_{z_0,w}}a_n(z-z_0)^n\\
\amp =\sum_{n=0}^\infty \frac{a_n}{(n+1)}((z-z_0)^{n+1})\Bigr\vert_{z_0}^w\\
\amp = \sum_{n=0}^\infty \frac{a_n}{n+1}(w-z_0)^{n+1}\\
\amp = \sum_{n=1}^\infty \frac{a_{n-1}}{n}(w-z_0)^{n}\text{.}
\end{align*}
It follows that
\(\sum_{n=1}^\infty \frac{a_{n-1}}{n}(w-z_0)^{n}\) converges, and thus so does
\(F(w)=C+\sum_{n=1}^\infty \frac{a_{n-1}}{n}(w-z_0)^{n}\text{.}\) We conclude that
\(F(z)\) converges for all
\(z\in B_R(z_0)\text{,}\) and hence that
\(R'\geq R\text{.}\) Since
\(F'(z)=f(z)\text{,}\) statement (3) of
Theorem 1.17.9 implies that
\(R\geq R'\text{.}\) We conclude that
\(R=R'\text{.}\)
Example 1.17.11. Differentiating power series.
Let \(f(z)=1/(1-z)^m\text{,}\) where \(m\) is a positive integer. Find a power series representation of \(f\) centered at \(z_0=0\) and compute its radius of convergence.
Solution.
The idea is to relate \(f\) with \(g(z)=1/(1-z)\text{.}\) Note that \(g'(z)=1/(1-z)^2\text{,}\) \(g''(z)=2/(1-z)^3\text{,}\) and more generally (using an easy induction argument)
\begin{equation*}
g^{(m)}(z)=m!(1-z)^{-(m+1)}=\frac{m!}{(1-z)^{m+1}}
\end{equation*}
for all \(m\geq 1\text{.}\) Thus we have
\begin{equation*}
f(z)=\frac{1}{(m-1)!}g^{(m-1)}(z)\text{.}
\end{equation*}
Now, beginning with the power series representation
\(g(z)=\sum_{n=0}^\infty z^n\) and using (3) of
Theorem 1.17.9, we see that
\begin{align*}
g'(z) \amp =\sum_{n=0}^\infty nz^{n-1}=\sum_{n=1}^\infty nz^{n-1}=\sum_{n=0}^\infty (n+1)a_{n+1}z^{n}\\
g''(z) \amp = \sum_{n=0}^\infty n(n-1)z^{n-2}=\sum_{n=0}^\infty (n+2)(n+1)z^{n}\\
\amp \vdots \ (\text{induction argument}) \\
g^{(m-1)}(z) \amp = \sum_{n=0}^\infty (n+(m-1))(n+(m-2))\cdots (n+1)z^{n}\\
\amp = \sum_{n=0}^\infty \frac{(n+m-1)!}{n!}z^{n} \text{.}
\end{align*}
We conclude that
\begin{align*}
\frac{1}{(1-z)^{m}} \amp = \frac{1}{(m-1)!}g^{(m-1)}(z)\\
\amp = \sum_{n=0}^\infty\frac{(n+m-1)!}{n!(m-1)!}z^{n}\\
\amp = \sum_{n=0}^\infty{n+m-1\choose n}z^{n}\text{,}
\end{align*}
where we use the binomial coefficient notation
\({r\choose s}=\tfrac{r!}{s!(r-s)!}\text{.}\) Note that since the original power series for
\(g(z)\) has radius of convergence
\(R=1\text{,}\) so too does each successive formal derivative of this power series. (See
Corollary 1.17.10.) Furthermore, since scaling a power series by a constant clearly does not change its radius of convergence, we conclude that our power series for
\(f(z)=1/(z-z_0)^m\) has radius of convergence
\(1\text{.}\)
Example 1.17.12. Power series for \(\Log\).
Find a power series expansion of \(-\Log(1-z)\) that converges on \(B_1(0)\text{.}\)
Solution.
Let \(f(z)=1/(1-z)\text{.}\) Since the power series expansion
\begin{equation*}
\frac{1}{1-z}=1+z+z^2+\cdots
\end{equation*}
has radius of convergence \(R=1\text{,}\) so too does the formal antiderivative
\begin{equation*}
F(z)=z+\frac{1}{2}z^2+\frac{1}{3}z^3+\cdots =\sum_{n=1}^\infty \frac{1}{n}z^n\text{,}
\end{equation*}
according to
Corollary 1.17.10. According to
Theorem 1.17.9, the power series
\(F(z)\) defines a holomorphic function on
\(B_1(0)\) and satisfies
\(F'(z)=f(z)\text{:}\) i.e., the function
\(F\) is an antiderivative of
\(f\) on
\(B_1(0)\text{.}\) Since
\(-\Log(1-z)\) is also an antiderivative of
\(f\) on
\(B_1(0)\text{,}\) as you can easily check, and since
\(B_1(0)\) is
connected, we have
\(-\Log(1-z)=F(z)+C\) for some
\(C\text{.}\) Evaluating at
\(z=0\text{,}\) we see that since
\(-\Log(1-0)=F(0)=0\text{,}\) we must have
\(C=0\text{.}\) We conclude that
\begin{equation*}
-\Log(1-z)=F(z)=\sum_{n=1}^\infty \frac{1}{n}z^{n}=z+\frac{z^2}{2}+\frac{z^3}{3}+\cdots \text{,}
\end{equation*}
and that his power series representation has radius of convergence \(1\text{.}\)
According to
Theorem 1.17.9, functions
\(f\) that admit a power series expansion
\(f(z)=\sum_{n=0}^\infty a_n(z-z_0)^n\) in some open ball around
\(z_0\) behave very nicely with respect to the operations of integration and differentiation. As such it is worth giving such functions a special designation:
analytic.
Definition 1.17.13. Analytic functions.
Let \(f\colon U\rightarrow \C\) be a complex function, with \(U\) an open set. We say \(f\) is analytic at a point \(z_0\in U\) if \(f\) has a power series representation \(f(z)=\sum_{n=0}^\infty a_n(z-z_0)^n\) with positive radius of convergence. We say \(f\) is analytic on a set \(D\subseteq U\) if \(f\) is analytic at all points in \(D\text{,}\) and we say \(f\) is analytic if it is analytic on its domain.
A direct consequence of
Theorem 1.17.9 is that analytic functions are holomorphic, and in fact infinitely differentiable. We make this official now.
Corollary 1.17.14. Analytic implies holomorphic.
Let \(f\) be a complex function. If \(f\) is analytic at a point \(z_0\in \C\text{,}\) then \(f\) is holomorphic at \(z_0\text{.}\) Using logical shorthand:
\begin{equation}
f \text{ analytic at } z_0 \implies f \text{ holomorphic at } z_0\tag{1.68}
\end{equation}
The next corollary provides a formula for the \(n\)-th coefficient of a power series expansion of \(f\) both in terms of the \(n\)-th derivative of \(f\) (as well as in term of certain line integrals). As a result, we see that a power series expansion of an analytic function is uniquely determined by the function.
Corollary 1.17.15. Uniqueness of expansions.
Let \(f\) be a complex function. If \(f\) is analytic at \(z_0\) with power series representation \(f(z)=\sum_{n=0}^\infty a_n(z-z_0)^n\text{,}\) then
\begin{equation}
a_n=\frac{1}{n!}f^{(n)}(z_0)=\frac{1}{2\pi i}\int_\gamma \frac{f(z)}{(z-z_0)^{n+1}}\text{,}\tag{1.69}
\end{equation}
where \(\gamma\) is the counterclockwise oriented boundary of any closed disk \(\overline{B}\subseteq U\) containing \(z_0\text{.}\)
In particular the power series expansion of \(f\) centered at \(z_0\) is unique.
Proof.
We prove the derivative formula for \(a_n\) for all \(n\geq 0\text{.}\) The integral formula follows directly from the generalized Cauchy integral formula.
We will prove that
\begin{equation}
f^{(n)}(z)=\sum_{k=0}^\infty \frac{(n+k)!}{k!}a_{k+n}(z-z_0)^k\text{,}\tag{1.70}
\end{equation}
and thus that \(f^{(n)}(z_0)=\frac{n!}{0!}a_n\text{,}\) or equivalently
\begin{equation*}
a_n=\frac{f^{(n)}(z_0)}{n!}\text{.}
\end{equation*}
The proof of formula
(1.70) is by induction. The base case (
\(n=0\)) is clear, since
\begin{equation*}
\sum_{k=0}^\infty \frac{(0+k)!}{k!}a_{k+0}(z-z_0)^k=\sum_{k=0}^\infty a_k(z-z_0)^k=f(z)\text{.}
\end{equation*}
Assuming now by induction that
(1.70) holds for
\(n\text{,}\) we have
\begin{align*}
f^{(n+1)}(z) \amp = \frac{d}{dz}(f^{(n)}(z))\\
\amp = \frac{d}{dz}\left(\sum_{k=0}^\infty \frac{(n+k)!}{k!}(z-z_0)^k\right) \amp (\text{induction hypo.}) \\
\amp = \sum_{k=0}^\infty k\cdot \frac{(n+k)!}{k!}(z-z_0)^{k-1} \amp (\knowl{./knowl/xref/th_power_series_integral_der.html}{\text{Theorem 1.17.9}})\\
\amp = \sum_{k=0}^\infty \frac{(n+k)!}{(k-1)!}(z-z_0)^{k-1} \\
\amp = \sum_{k=1}^\infty \frac{(n+k)!}{(k-1)!}(z-z_0)^{k-1} \\
\amp = \sum_{k=0}^\infty \frac{(n+1+k)!}{k!}(z-z_0)^{k} \text{,}
\end{align*}
as desired.
As always in mathematics, when confronted with an implication like
(1.68), we should always ask whether its converse is true: i.e., whether the implication is in fact an equivalence. In the next section we will see that this is indeed the case for
(1.68): i.e., we have
\begin{equation}
f \text{ analytic at } z_0 \iff f \text{ holomorphic at } z_0\text{.}\tag{1.71}
\end{equation}
