Section 7.5 Applications of Taylor and Laurent Series
In this section we show how you can use Taylor and Laurent series to derive important properties of analytic functions. We begin by showing that the zeros of an analytic function must be isolated unless the function is identically zero. A point \(\alpha\) of a set \(T\) is called isolated if there exists a disk \(D_{R}(\alpha)\) about \(\alpha\) that does not contain any other points of \(T\text{.}\)
Theorem 7.5.1.
Suppose that \(f\) is analytic in a domain \(D\) containing \(\alpha\) and that \(f(\alpha) =0\text{.}\) If \(f\) is not identically zero in \(D\text{,}\) then there exists a punctured disk \(D_{R}^*(\alpha)\) in which \(f\) has no zeros.
Proof.
By Taylor’s theorem, there exists some disk \(D_{R}(\alpha)\) about \(\alpha\) such that
\begin{equation*}
f(z) = \sum_{n=0}^{\infty}\frac{f\,^{(n)}(\alpha)}{n!}(z-\alpha)^n \text{ for all } z \in D_{R}(\alpha)\text{.}
\end{equation*}
If all the Taylor coefficients
\(\frac{f\,^{(n)}(\alpha)}{n!}\) of
\(f\) were zero, then
\(f\) would be identically zero on
\(D_{R}(\alpha)\text{.}\) A proof similar to the proof of the maximum modulus principle (
Theorem 6.6.3) would then show that
\(f\) is identically zero in
\(D\text{,}\) contradicting our assumption about
\(f\text{.}\)
Thus, not all the Taylor coefficients of
\(f\) are zero, and we may select the smallest integer
\(k\) such that
\(\frac{f\,^{(k)}(\alpha)}{k!} \ne 0\text{.}\) According to the results in
Section 7.4,
\(f\) has a zero of order
\(k\) at
\(\alpha\) and can be written in the form
\begin{equation*}
f(z) = (z-\alpha)^kg(z)\text{,}
\end{equation*}
where \(g\) is analytic at \(\alpha\) and \(g(\alpha) \ne 0\text{.}\) Since \(g\) is a continuous function, there exists a disk \(D_r(\alpha)\) throughout which \(g\) is nonzero. Therefore \(f(z) \ne 0\) in the punctured disk \(D_r^{\ast }(\alpha)\text{.}\)
The proofs of the following two corollaries are given as exercises.
Corollary 7.5.2.
Suppose that \(f\) is analytic in the domain \(D\) and that \(\alpha \in D\text{.}\) If there exists a sequence of points \(\{z_n\} \in D\) such that \(z_n \to \alpha\text{,}\) and \(f(z_n) =0\text{,}\) then \(f(z)=0\) for all \(z \in D\text{.}\)
Corollary 7.5.3.
Suppose that \(f\) and \(g\) are analytic in the domain \(D\text{,}\) where \(\alpha \in D\text{.}\) If there exists a sequence \(\{z_n\} \in D\) such that \(z_n \to \alpha\text{,}\) and \(f(z_n)=g(z_n)\) for all \(n\text{,}\) then \(f(z)=g(z)\) for all \(z \in D\text{.}\)
Theorem 7.5.1 also allows us to give a simple argument for one version of L’Hôpital’s rule.
Corollary 7.5.4.
(L’Hôpital’s rule) Suppose that \(f\) and \(g\) are analytic at \(\alpha\text{.}\) If \(f(\alpha)=0\) and \(g(\alpha)=0\text{,}\) but \(g\,'(\alpha) \ne 0\text{,}\) then
\begin{equation*}
\lim_{z \to \alpha }\frac{f(z)}{g(z)} = \frac{f\,'(\alpha)}{g'(\alpha)}
\end{equation*}
Proof.
Because
\(g\,'(\alpha) \ne0\text{,}\) \(g\) is not identically zero and, by
Theorem 7.5.1, there is a punctured disk
\(D_r^*(\alpha)\) in which
\(g(z) \ne 0\text{.}\) Thus the quotient
\(\frac{f(z)}{g(z)}=\frac{f(z) -f(\alpha)}{g(z) -g(\alpha)}\) is defined for all
\(z \in D_r^*(\alpha)\text{,}\) and we can write
\begin{equation*}
\lim_{z \to\alpha}\frac{f(z)}{g(z)} = \lim_{z \to\alpha}\frac{f(z) -f(\alpha)}{g(z) - g(\alpha)} = \lim_{z \to\alpha}\frac{[f(z) - f(\alpha)] / (z-\alpha)}{[g(z) - g(\alpha)] /(z-\alpha)} = \frac{f\,'(\alpha)}{g'(\alpha)}\text{.}
\end{equation*}
We can use the following Theorem to get Taylor series for quotients of analytic functions. Its proof involves ideas from
Section 7.2, and we leave it as an exercise.
Theorem 7.5.5. Division of power series.
Suppose that \(f\) and \(g\) are analytic at \(\alpha\) with the power series representations
\begin{equation*}
f(z) = \sum_{n=0}^{\infty}a_n(z-\alpha)^n \text{ and } g(z) = \sum_{n=0}^{\infty}b_n(z-\alpha)^n, \text{ for all } z \in D_{R}(\alpha)\text{.}
\end{equation*}
If \(g(\alpha) \ne 0\text{,}\) then the quotient \(\frac{f}{g}\) has the power series representation
\begin{equation*}
\frac{f(z)}{g(z)} = \sum_{n=0}^{\infty}c_n(z-\alpha)^n\text{,}
\end{equation*}
where the coefficients satisfy the equation
\begin{equation*}
a_n = b_0c_n + \cdots + b_{n-1}c_1 + b_nc_0\text{.}
\end{equation*}
In other words, we can obtain the series for the quotient \(\frac{f(z)}{g(z)}\) by the familiar process of dividing the series for \(f(z)\) by the series for \(g(z)\text{,}\) using the standard long division algorithm.
Example 7.5.6.
Find the first few terms of the Maclaurin series for the function \(f(z) =\sec z\text{,}\) if \(|z| \lt \frac{\pi}{2}\text{,}\) and compute \(f\,^{(4)}(0)\text{.}\)
Solution.
Using long division, we see that
\begin{equation*}
\sec z = \frac{1}{\cos z} = \frac{1}{1-\frac{z^2}{2!}+\frac{z^4}{4!}-\frac{z^{6}}{6!}+\cdots } = 1+\frac{1}{2}z^2+\frac{5}{24}z^4+\cdots\text{.}
\end{equation*}
Using Taylor’s theorem, we see that if \(f(z) =\sec z\text{,}\) then \(\frac{f\,^{(4)}(0)}{4!}=\frac{5}{24}\text{,}\) so that \(f\,^{(4)}(0) = 5\text{.}\)
We close this section with some results concerning the behavior of complex functions at points near the different types of isolated singularities. The following is due to the German mathematician G. F. Bernhard Riemann (1826–1866).
Theorem 7.5.7. Riemann.
Suppose that \(f\) is analytic in \(D_r^*(\alpha)\text{.}\) If \(f\) is bounded in \(D_r^*(\alpha)\text{,}\) then either \(f\) is analytic at \(\alpha\) or \(f\) has a removable singularity at \(\alpha\text{.}\)
Proof.
Consider the function \(g\text{,}\) defined as
\begin{equation}
g(z) = \begin{cases} (z-\alpha)^2f(z) , \amp \text{ when } z \ne \alpha ; \\ \qquad 0, \amp \text{ when } z=\alpha. \end{cases}\tag{7.5.1}
\end{equation}
Clearly, \(g\) is analytic in at least \(D_r^*(\alpha)\text{.}\) Straightforward calculation yields
\begin{equation*}
g\,'(\alpha) = \lim_{z \to\alpha}\frac{g(z) -g(\alpha)}{z-\alpha}=\lim_{z \to\alpha }(z-\alpha)f(z) = 0\text{.}
\end{equation*}
The last equation follows because \(f\) is bounded. Thus \(g\) is also analytic at \(\alpha\text{,}\) with \(g(\alpha) = g\,'(\alpha) = 0\text{.}\)
By Taylor’s theorem, \(g\) has the representation
\begin{equation}
g(z) = \sum_{n=2}^{\infty}\frac{g^{(n)}(\alpha)}{n!}(z-\alpha)^n, \text{ for all } z \in D_r(\alpha)\text{.}\tag{7.5.2}
\end{equation}
We divide both sides of Equation
(7.5.2) by
\((z-\alpha)^2\) and use Equation
(7.5.1) to obtain the following power series representation for
\(f\text{:}\)
\begin{equation*}
f(z) = \sum_{n=2}^{\infty}\frac{g^{(n)}(\alpha)}{n!}(z-\alpha)^{n-2} = \sum \limits_{n=0}^{\infty}\frac{g^{(n+2)}(\alpha)}{(n+2)!}(z-\alpha)^n\text{.}
\end{equation*}
By
Theorem 4.4.7,
\(f\) is analytic at
\(\alpha\) if we define
\(f(\alpha)=\frac{g^{(2)}(\alpha)}{2!}\text{.}\) This observation completes the proof.
The proof of the following Corollary is given as an exercise.
Corollary 7.5.8.
If \(f\) is analytic in \(D_r^*(\alpha)\text{,}\) then \(f\) can be defined to be analytic at \(\alpha\) iff \(\lim\limits_{z \to \alpha }f(z)\) exists and is finite.
Theorem 7.5.9.
Suppose that \(f\) is analytic in \(D_r^*(\alpha)\text{.}\) The function \(f\) has a pole of order \(k\) at \(\alpha\text{,}\) iff \(\lim\limits_{z \to\alpha}|f(z)| = \infty\text{.}\)
Proof.
Suppose, first, that
\(f\) has a pole of order
\(k\) at
\(\alpha\text{.}\) Using
Theorem 7.4.8, we can say that
\(f(z) = \frac{h(z)}{(z-\alpha)^k}\text{,}\) where
\(h\) is analytic at
\(\alpha\text{,}\) and
\(h(\alpha) \ne 0\text{.}\) Because
\(\lim\limits_{z \to a}|h(z)| =|h(\alpha)| \ne 0\) and
\(\lim\limits_{z \to a}|(z-\alpha)|=0\text{,}\) we conclude that
\(\lim\limits_{z \to a}|f(z)| = \lim\limits_{z \to a}|h(z)| \lim\limits_{z \to a}\frac{1}{|(z-\alpha)^k| } = \infty\text{.}\)
Conversely, suppose that
\(\lim\limits_{z \to \alpha }|f(z)| = \infty\text{.}\) By the definition of a limit, there must be some
\(\delta >0\) such that
\(|f(z)| > 1\) if
\(z \in D_{\delta }^{\ast }(\alpha)\text{.}\) Thus the function
\(g(z) = \frac{1}{f(z)}\) is analytic and bounded (because
\(|g(z)| =|\frac{1}{f(z)}| \le 1\)) in
\(D_{\delta }^{\ast }(\alpha)\text{.}\) By
Theorem 7.5.7, we may define
\(g\) at
\(\alpha\) so that
\(g\) is analytic in all of
\(D_{\delta }(\alpha)\text{.}\) In fact,
\(|g(\alpha)| = \lim\limits_{z \to a}\frac{1}{|f(z)|}=0\text{,}\) so
\(\alpha\) is a zero of
\(g\text{.}\) We claim that
\(\alpha\) must be of finite order; otherwise, we would have
\(g\,^{(n)}(\alpha)=0\text{,}\) for all
\(n\text{,}\) and hence
\(g(z) = \sum\limits_{n=0}^{\infty}\frac{g\,^{(n)}(\alpha}{n!}(z-\alpha)^n=0\text{,}\) for all
\(z \in D_{\delta}(\alpha)\text{.}\) As
\(g(z) = \frac{1}{f(z)}\) is analytic in
\(D_{\delta }^{\ast }(\alpha)\text{,}\) this result is impossible, so we can let
\(k\) be the order of the zero of
\(g\) at
\(\alpha\text{.}\) By
Corollary 7.4.9,
\(f\) has a pole of order
\(k\text{.}\)
Theorem 7.5.10.
The function \(f\) has an essential singularity at \(\alpha\) iff \(\lim\limits_{z \to\alpha}|f(z) |\) does not exist.
Proof.
Example 7.5.11.
Show that the function \(g\) defined by
\begin{equation*}
g(z) = \begin{cases}e^{-\left(\frac{1}{z^2}\right)}, \amp \ \ \text{ when } z \ne 0\text{ , and } \\ 0, \amp \text{ when } z=0, \end{cases}
\end{equation*}
is not continuous at \(z=0\text{.}\)
Solution.
In Exercise 20,
Section 7.2, we asked you to show this relation by computing limits along the real and imaginary axes. Note, however, that the Laurent series for
\(g(z)\) in the annulus
\(D_r^{\ast }(0)\) is
\begin{equation*}
g(z) = 1+\sum\limits_{n=1}^{\infty}(-1)^n\frac{1}{z^{2n}}\text{,}
\end{equation*}
so that
\(0\) is an essential singularity for
\(g\text{.}\) According to
Theorem 7.5.10,
\(\lim\limits_{z \to 0}|g(z)|\) doesn’t exist, so
\(g\) is not continuous at
\(0\text{.}\)
Exercises Exercises
1.
Determine whether there exists a function \(f\) that is analytic at \(0\) with the property that, for \(n=1,\,2,\,3,\ldots\text{,}\)
(a)
\(f(\frac{1}{2n}) = 0\) and \(f(\frac{1}{2n-1}) = 1\text{.}\)
Solution.
No. Otherwise \(0=\lim\limits_{n \to \infty}f(\frac{1}{2n}) = f(\lim\limits_{n \to \infty} \frac{1}{2n}) = f(0)\text{.}\) On the other hand, \(1=\lim\limits_{n \to \infty}f(\frac{1}{2n-1}) = f\left(\lim\limits_{n \to \infty}\frac{1}{2n-1}\right) = f(0)\text{.}\) Justify and explain.
(b)
\(f(\frac{1}{n}) =f(-\frac{1}{n}) = \frac{1}{n^2}\text{.}\)
Solution.
Yes. There is a simple function with this property. Find it.
(c)
\(f(\frac{1}{n}) =f(-\frac{1}{n}) = \frac{1}{n^3}\text{.}\)
Solution.
No. Use
Corollary 7.5.3 to show that for all
\(z\) in some disk
\(D_{r}(0)\) we have
\(f(z)=z^3\text{,}\) and
\(f(z) = -z^3\text{,}\) and explain why this is impossible.
2.
Prove the following results.
(a)
(b)
(c)
(d)
3.
Consider the function \(f(z) =z\sin (\frac{1}{z})\text{.}\)
(a)
Show that there is a sequence \(\{z_n\}\) of points converging to \(0\) such that \(f(z_n) =0\) for \(n=1,\,2,\,3,\ldots\) .
Solution.
Let \(z_n=\frac{1}{\pi n}\text{.}\) Explain.
(b)
Solution.
No, the function \(f\) is not analytic at zero (explain why), which is required by the corollary.
4.
Let \(f(z) =\tan z\text{.}\)
(a)
Use
Theorem 7.5.5 to find the first few terms of the Maclaurin series for
\(f(z)\text{,}\) if
\(|z| \lt \frac{\pi}{2}\text{.}\)
(b)
What are the values of \(f\,^{(6)}(0)\) and \(f\,^{(7)}(0)\text{?}\)
5.
Show that the real function \(f\) defined by
\begin{equation*}
f(x) = \begin{cases}x\sin (\frac{1}{x}) , \amp \text{ when } x \ne 0\text{ , and } \\ 0, \amp \text{ when } x=0, \end{cases}
\end{equation*}
is continuous at \(x=0\) but that the corresponding function \(g(z)\) defined by
\begin{equation*}
g(z) = \begin{cases}z\sin (\frac{1}{z}) , \amp \text{ when } z \ne 0, \text{ and } \\ 0, \amp \text{ when } z=0, \end{cases}
\end{equation*}
is not continuous at \(z=0\text{.}\)
Solution.
For
\(x\ne 0, \, \lim\limits_{x \to 0}|x\sin \frac{1}{x}| \le \lim\limits_{x\to 0}|x|=0\text{.}\) This implies
\(\lim\limits_{x\to 0}f(x) =0 = f(0)\text{.}\) For the complex case, show that there is an essential singularity at 0 and use
Theorem 7.5.10.
6.
Use L’Hôpital’s rule to find the following limits.
(a)
\(\lim\limits_{z \to 1+i}\frac{z-1-i}{z^4+4}\text{.}\)
(b)
\(\lim\limits_{z \to i}\frac{z^2-2iz-1}{z^4+2z^2+1}\text{.}\)
(c)
\(\lim\limits_{z \to i}\frac{1+z^{6}}{1+z^2}\text{.}\)
(d)
\(\lim\limits_{z \to 0}\frac{\sin z+\sinh z-2z}{z^{5}}\text{.}\)
