Lecture 10: Taylor Series for Complex Functions

In this lecture, we reproduce the theory of Taylor expansion in calculus to analytic functions on the complex plane.

General power series

We start with the general power series of the form $\label{EQ:Power Series} w(z)=\displaystyle\sum_{j=0}^\infty c_j (z-z_0)^j .$ We call the constants $\{c_j\}_{j=0}^\infty$ the coefficients of the power series. This power series is simply a translation of the power series $\label{EQ:Power Series 0} w(z)=\displaystyle\sum_{j=0}^\infty c_j z^j .$ It is therefore sufficient that we study the properties of this power series.

One of the most important concepts in the study of power series is the radius of convergence of a series.

Definition 1. The largest number $R$ such that the power series [EQ:Power Series 0] converges for all $|z|<R$ is called the radius of convergence of the power series. The set $\{ z\in\mathbb C\mid |z|=R\}$ is called the circle of convergence for this power series.

A rigorous treatment of radius of convergence requires the concepts of limit superior and limit inferior. Here we provide the following result without proof.

Theorem 2. Let $R$ be the radius of convergence for [EQ:Power Series 0]. Then $\dfrac{1}{R} = \lim_{j\to\infty} \left|\dfrac{c_{j+1}}{c_j}\right|$ provided that the limit exists.

An illustration of the convergence property of the power series [EQ:Power Series] is provided in Figure 1.

Disk centered at z0 with radius R; the series converges inside and diverges outside. — Illustration of convergence and divergence regions of the the power series [EQ:Power Series].

Example 3. The series $\sum_{j=0}^\infty z^j$ has radius of convergence $R=1$ .

Example 4. The series $\sum_{j=0}^\infty \dfrac{z^j}{j!}$ has radius of convergence $R=\infty$ . This is because $\lim_{j\to \infty} |\dfrac{1/(j+1)!}{1/j!}|=\lim_{j\to\infty}\dfrac{1}{j+1} =0$ .

Example 5. The series $\sum_{j=0}^\infty \dfrac{j!}{j^j} z^j$ has radius of convergence $R=e$ . This is because $\lim_{j\to \infty} \dfrac{(j+1)!}{(j+1)^{j+1}}/\dfrac{j!}{j^j}=\lim_{j\to\infty}\left(\dfrac{j}{j+1}\right)^j =\dfrac{1}{e}$ .

One fact that makes the power series [EQ:Power Series 0] (or in general [EQ:Power Series]) special is that each term in the power series is an analytic function (since it’s an integer power of $z$ (or in general $z-z_0$ )). This means that the power series is a summation of analytic functions. Therefore, if the power series converges, we can expect that the sum is analytic.

Theorem 6. The power series [EQ:Power Series 0] sums to a function that is analytic at each point inside of its circle of convergence.

Proof. Let $s_n(z)$ be the sequence of partial sum. This sequence converges uniformly to $f(z)$ inside the circle of convergence. Therefore $\int_C s_n(z) dz \to \int_C f(z) dz$ for any contour inside the circle of convergence. Due to analyticity of $s_n(z)$ , $\int_C s_n(z) dz=0$ . Therefore $\int_C f(z) dz=0$ for all $C$ inside the circle of convergence. Therefore, $f(z)$ has an anti-derivative that is analytic inside the circle by Morera’s Theorem. ◻

Taylor series

The following theorem states that one can write an analytic function as a power series within its region of analyticity.

Theorem 7 (Taylor Series). Let $f(z)$ be analytic in $\{ z\in\mathbb C\mid |z-z_0|\le R\}$ for some $R>0$ . Then $\label{EQ:Taylor} f(z)=\sum_{k=0}^\infty \dfrac{f^{(k)}(z_0)}{k!}(z-z_0)^k$ converges uniformly in $\{ z\in\mathbb C\mid |z-z_0|\le \widehat R\}$ for any $\widehat R<R$ .

Proof. By the Cauchy’s Integral Formula, we have, $\forall z$ s.t. $|z-z_0|<R$ , $\begin{gathered} f(z)=\dfrac{1}{2\pi i}\int_{|\zeta-z_0|=R} \dfrac{f(\zeta)}{\zeta-z} d\zeta \\ =\dfrac{1}{2\pi i}\int_{|\zeta-z_0|=R} \dfrac{f(\zeta)}{\zeta-z_0}\dfrac{1}{1-\dfrac{z-z_0}{\zeta-z_0}} d\zeta=\dfrac{1}{2\pi i}\int_{|\zeta|=R} \dfrac{f(\zeta+z_0)}{\zeta}\dfrac{1}{1-\dfrac{z-z_0}{\zeta}} d\zeta. \end{gathered}$ Using the fact that $\dfrac{1}{1-z/\zeta}=\sum_{k=0}^\infty (z/\zeta)^k$ converges when $|z|<|\zeta|$ , we can have $f(z)=\dfrac{1}{2\pi i}\int_{|\zeta|=R} \dfrac{f(\zeta+z_0)}{\zeta} \sum_{k=0}^\infty (\frac{z-z_0}{\zeta})^k d\zeta.$ When $|z|\le \widehat R<R$ , the series inside converges uniformly. We can therefore exchange the order of taking limit and performing integration. We therefore have $f(z)=\dfrac{1}{2\pi i}\sum_{k=0}^\infty (z-z_0)^k \int_{|\zeta|=R} \dfrac{f(\zeta+z_0)}{\zeta^{k+1}} d\zeta=\sum_{k=0}^\infty \dfrac{f^{(k)}(z_0)}{k!}(z-z_0)^k.$ We have used Cauchy’s Derivative Formula in the last step. ◻

Definition 8 (Taylor Series). We call the series given in [EQ:Taylor] the Taylor series of $f(z)$ at $z=z_0$ .

This definition has exactly the same format as Taylor series in the real variable case. In fact, for elementary functions, we can show that their Taylor series have exactly the same forms as their real-variable counterparts. Taylor series is an important tool in studying analytic functions.

Example 9. Let $f(z)=e^z$ . Then $f(z)$ is analytic on $\mathbb C$ . Moreover, $f^{(k)}(z)=e^z$ . Therefore the Taylor series for $f(z)$ at $z_0$ is $e^z = \sum_{k=0}^\infty \dfrac{e^{z_0}}{k!}(z-z_0)^k =e^{z_0} \sum_{k=0}^\infty \dfrac{(z-z_0)^k}{k!}$ When $z_0=0$ , this simplifies to $e^z = \sum_{k=0}^\infty \dfrac{ z^k}{k!}$

Example 10. Let $f(z)=\sin z$ . Then $f(z)$ is analytic on $\mathbb C$ . Moreover, $f^{(k)}(z)=(-1)^{[k/2]} \left \{ \begin{array}{cc} \cos z, & k \ \mbox{odd}\\ \sin z, & k\ \mbox{even} \end{array}\right.\ k\ge 1.$ To consider the Taylor series of $\sin z$ at $z_0=0$ , we observe that $\sin 0=0$ and $\cos 0=1$ . Therefore, the even terms ( $0,\ 2,\ 4,\ \cdots$ ) all disappear in the series. The odd terms have alternative signs since $(-1)^{[k/2]}$ alternates between $1$ (when $k=4m+1$ and $-1$ otherwise. Therefore the Taylor series for $f(z)$ at $z_0=0$ is $\sin z = \sum_{k=1}^\infty (-1)^{[k-1/2]} (z)^{2k-1} =z-\dfrac{z^3}{3!}+\dfrac{z^5}{5!}-\cdots$

Theorem Theorem 7 and a straightforward formal calculation lead to the following result.

Theorem 11 (Term by Term Differentiation). Let $f(z)$ be analytic in $\{z\in\mathbb C\mid |z-z_0|\le R\}$ for some $R>0$ . Then the series obtained by differentiating the Taylor series of $f(z)$ termwise converges uniformly to $f'(z)$ in $\{z\in\mathbb C\mid |z-z_0|\le \widehat R\}$ for any $\widehat R<R$ .

Proof. By the Cauchy’s Derivative Formula, we have, $\forall z$ s.t. $|z-z_0|=\widehat R<R$ , $f'(z)=\dfrac{1}{2\pi i}\int_{|\zeta-z_0|=R} \dfrac{f(\zeta)}{(\zeta-z)^2} d\zeta=\dfrac{1}{2\pi i}\int_{|\zeta-z_0|=R} \dfrac{\sum_{k=0}^\infty c_k (\zeta-z_0)^k}{(\zeta-z)^2} d\zeta,\ \ c_k=\dfrac{f^{(k)}(z_0)}{k!}\,.$ When $|z|\le \widehat R<R$ , the series inside converges uniformly. We can therefore exchange the order of taking limit and performing integration. We therefore have $f'(z)=\sum_{k=0}^\infty c_k \left(\dfrac{1}{2\pi i}\int_{|\zeta-z_0|=R} \dfrac{(\zeta-z_0)^k}{(\zeta-z)^2} d\zeta\right)=\sum_{k=1}^\infty c_k k (z-z_0)^{k-1}=\sum_{k=1}^\infty \dfrac{f^{(k)}(z_0)}{(k-1)!}(z-z_0)^{k-1},$ where we have used Cauchy’s Derivative Formula in the last step.

This formula is exactly what we would get by differentiating the Taylor series of $f(z)$ term by term. ◻

Guided review and additional examples

Learning goals

Compute Taylor coefficients from derivatives.
Recognize standard power series and shift their centers.
Determine a radius of convergence from nearby singularities.
Differentiate and integrate power series inside their convergence disks.

Concept connection: geometry determines convergence

For an analytic function expanded at $z_0$ , the Taylor series converges at least until it reaches the nearest singularity of the function’s analytic continuation. The radius is therefore a geometric distance in the complex plane, even when the original question is posed for real values.

Worked example 1: shift the center of the exponential series

Because every derivative of $e^z$ is $e^z$ , expansion about $z_0=1$ gives

$e^z=e\sum_{n=0}^{\infty}\frac{(z-1)^n}{n!}.$

The exponential has no finite singularities, so the radius of convergence is infinite.

Worked example 2: locate the nearest singularities

The function $f(z)=1/(1+z^2)$ has singularities at $z=i$ and $z=-i$ . Both are distance $1$ from the origin, so its Taylor series about $0$ has radius $R=1$ . Indeed,

$\frac1{1+z^2}=\frac1{1-(-z^2)} =\sum_{n=0}^{\infty}(-1)^n z^{2n},\qquad |z|<1.$

Check your understanding

Why does the geometric series $1/(1-z)=\sum_{n=0}^{\infty}z^n$ stop converging at radius $1$ ?

Show the answer

The function has a singularity at $z=1$ , exactly distance $1$ from the center $0$ . The terms also fail to approach zero when $|z|\ge1$ , confirming that the disk $|z|<1$ is maximal.