Analytic Functions

From Holomorphic Functions to Power Series

The Cauchy–Riemann Equations showed how real partial derivatives can establish complex differentiability. A power series gives a different kind of description: it represents a function locally by a sum of powers of the complex variable. Such representations are especially useful because their convergence behavior can be studied, and within their disks of convergence they can be differentiated term by term.

In this tutorial, we define analytic functions using local power series and prove that every analytic function is holomorphic. The direction matters: the results established here do not yet prove that every holomorphic function has a local power series representation. We will also prove that the coefficients of a representation at a fixed center are uniquely determined by the function.

Definition (Analytic at a Point): Let $U\subseteq\mathbb{C}$ be open and $f:U\to\mathbb{C}$. The function $f$ is analytic at $z_0\in U$ if there are a number $r>0$ and complex coefficients $a_0,a_1,\ldots$ such that the disk $\{z:|z-z_0|<r\}$ is contained in $U$, the power series $\sum_{n=0}^{\infty}a_n(z-z_0)^n$ converges there, and $$ f(z)=\sum_{n=0}^{\infty}a_n(z-z_0)^n $$ for every $z$ in that disk. The function is analytic on $U$ if it is analytic at every point of $U$.

The center $z_0$ and the size of the disk may depend on the point. A series centered at one point can sometimes be re-expanded around another, but the definition only requires a representation on some neighborhood of each point. The radius of convergence of a particular series need not be the largest disk on which the function can be represented by some other series.

Power Series Are Holomorphic Inside Their Radius

The Cauchy–Hadamard Formula and Convergence Regions established that a power series converges absolutely inside its radius of convergence. The Uniform Convergence Inside the Radius theorem gives uniform convergence on each smaller closed disk. We now show that the sum is complex differentiable at every point strictly inside the radius, and identify its derivative.

Theorem (Differentiation of a Power Series): Suppose $f(z)=\sum_{n=0}^{\infty}a_n(z-z_0)^n$ has radius of convergence $R>0$, where $R$ may be infinite. Then $f$ is holomorphic on $\{z:|z-z_0|<R\}$, and for every $w$ in that disk, $$ f'(w)=\sum_{n=1}^{\infty}n a_n(w-z_0)^{n-1}. $$ The derivative series also converges absolutely at every point of the disk.

Proof. Fix $w$ with $|w-z_0|<R$, and write $q=w-z_0$. Choose $\delta>0$ such that $|q|+\delta<R$; if $R$ is infinite, any positive $\delta$ will do. For $|h|<\delta$, expand each term by the binomial theorem:

(q+h)^n=\sum_{k=0}^{n}\binom{n}{k}q^{n-k}h^k.

The sum of the absolute values of all terms in the resulting double series is bounded by

\sum_{n=0}^{\infty}|a_n| \sum_{k=0}^{n}\binom{n}{k}|q|^{n-k}|h|^k = \sum_{n=0}^{\infty}|a_n|(|q|+|h|)^n.

This last series converges because $|q|+|h|<R$. Therefore the double series is absolutely convergent and may be regrouped by powers of $h$. Define

b_k=\sum_{n=k}^{\infty}a_n\binom{n}{k}q^{n-k}.

These coefficients are finite: for any fixed $s$ with $|q|<s<R$, the same absolute-convergence argument with a positive increment smaller than $s-|q|$ bounds the sum defining each $b_k$. Regrouping gives a power series around $w$:

f(w+h)=\sum_{k=0}^{\infty}b_k h^k \qquad (|h|<\delta).

In particular, $b_0=f(w)$. For nonzero $h$, subtract $b_0$ and divide by $h$:

\frac{f(w+h)-f(w)}{h} =b_1+\sum_{k=2}^{\infty}b_k h^{k-1}.

Choose $0<\rho<\delta$. Since the power series at $w$ converges absolutely at $h=\rho$, the quantity $\sum_{k=2}^{\infty}|b_k|\rho^{k-2}$ is finite. For $0<|h|<\rho$, it follows that

\left|\sum_{k=2}^{\infty}b_k h^{k-1}\right| \leq |h|\sum_{k=2}^{\infty}|b_k|\rho^{k-2} \longrightarrow 0 \qquad\text{as }h\to0.

Thus $f'(w)=b_1$. From the definition of $b_k$,

b_1=\sum_{n=1}^{\infty}n a_n q^{n-1} =\sum_{n=1}^{\infty}n a_n(w-z_0)^{n-1}.

This sum is absolutely convergent: the absolute-convergence argument above bounds the binomial expansion for any sufficiently small positive increment, and its terms with $k=1$ have finite sum. Since $w$ was arbitrary, $f$ is complex differentiable throughout the disk, hence holomorphic there. $\square$

The proof does more than take a formal derivative of each term. It first produces a power series centered at an arbitrary interior point $w$. The coefficient of its linear term is precisely the complex derivative there. This local re-expansion is what justifies term-by-term differentiation.

Worked Examples

Worked Example: The Exponential Function

Consider the power series

E(z)=\sum_{n=0}^{\infty}\frac{z^n}{n!}.

For $z\neq0$, the ratio of the magnitudes of consecutive terms is

\frac{|z|^{n+1}/(n+1)!}{|z|^n/n!} =\frac{|z|}{n+1} \longrightarrow 0.

Thus the series converges for every $z\in\mathbb{C}$, so its radius of convergence is infinite. By the Differentiation of a Power Series theorem, $E$ is holomorphic everywhere, and

E'(z)=\sum_{n=1}^{\infty}\frac{n z^{n-1}}{n!} =\sum_{n=1}^{\infty}\frac{z^{n-1}}{(n-1)!} =E(z).

In the last equality, the index $m=n-1$ changes the sum into $\sum_{m=0}^{\infty}z^m/m!$. The function defined by this series is commonly denoted $e^z$.

Worked Example: A Geometric Series for a Rational Function

For $|z|<2$, the geometric series with ratio $z/2$ gives

\frac{1}{2-z} =\frac{1}{2}\frac{1}{1-z/2} =\sum_{n=0}^{\infty}\frac{z^n}{2^{n+1}}.

The ratio has modulus less than $1$ exactly when $|z|<2$, so this representation has radius of convergence $2$. Differentiating inside that disk yields

\left(\frac{1}{2-z}\right)' =\sum_{n=1}^{\infty}\frac{n z^{n-1}}{2^{n+1}}.

The derivative is also directly calculated from the rational expression as $(2-z)^{-2}$. At the center, the series derivative gives $1/4$: only its $n=1$ term is nonzero at $z=0$. Direct differentiation gives $(2-0)^{-2}=1/4$, in agreement.

Worked Example: Re-Expanding Around a Different Center

The function $g(z)=1/(z+1)$ has a power series around $z_0=1$. Write $z+1=2+(z-1)$. When $|z-1|<2$, the geometric series gives

g(z)=\frac{1}{2+(z-1)} =\frac{1}{2}\frac{1}{1+(z-1)/2} =\sum_{n=0}^{\infty}\frac{(-1)^n(z-1)^n}{2^{n+1}}.

This representation is centered at $1$, not at $0$, and converges on the disk $|z-1|<2$. Its derivative at $z=1$ is the coefficient of $z-1$, namely $-1/4$. Direct differentiation gives $g'(z)=-1/(z+1)^2$, so $g'(1)=-1/4$. This illustrates why analyticity is a local condition: a useful series can have a center chosen to suit the point of interest.

Coefficients of a Local Representation Are Unique

A function might initially appear to have more than one power series representation around the same point. The next theorem rules this out: once the center is fixed, the coefficients are determined by the function on any neighborhood where the representations agree.

Theorem (Uniqueness of Power Series Coefficients): Suppose two power series centered at $z_0$ both converge on some disk of positive radius about $z_0$ and represent the same function on a possibly smaller disk. Then their coefficients agree term by term.

Proof. Let the coefficients be $a_n$ and $c_n$, and set $d_n=a_n-c_n$. On a disk about $z_0$, equality of the represented functions gives

\sum_{n=0}^{\infty}d_n(z-z_0)^n=0.

Suppose, for contradiction, that some $d_n$ is nonzero. Since the nonnegative integers are well ordered, there is a least index $m$ such that $d_m\neq0$. For $z\neq z_0$, divide the displayed identity by $(z-z_0)^m$:

0=d_m+\sum_{n=m+1}^{\infty}d_n(z-z_0)^{n-m}.

The tail tends to zero as $z\to z_0$. To see this, choose $r>0$ strictly smaller than the common disk radius. The series $\sum_{n=m+1}^{\infty}|d_n|r^{n-m-1}$ is finite by absolute convergence inside the radius. For $|z-z_0|<r$, the tail has magnitude at most

|z-z_0|\sum_{n=m+1}^{\infty}|d_n|r^{n-m-1},

which tends to zero. Taking the limit as $z\to z_0$ in the divided identity gives $d_m=0$, contradicting the choice of $m$. Hence every $d_n=0$, so $a_n=c_n$ for all $n$. $\square$

Together with the differentiation theorem, uniqueness also identifies coefficients through derivatives. For a series centered at $z_0$, the constant coefficient is $f(z_0)$, and the linear coefficient is $f'(z_0)$. Repeated differentiation gives the familiar higher-order coefficient formulas whenever the successive derivatives are taken. The essential point here is that such coefficients cannot be changed while leaving the represented function unchanged near the center.

Analyticity and Holomorphicity

By definition, an analytic function has a local power series representation at every point. The Differentiation of a Power Series theorem therefore implies that it is holomorphic at every point. In particular, the Cauchy–Riemann Equations as Necessary Conditions apply to its real and imaginary parts wherever their partial derivatives are considered.

The proof established one implication: analytic functions are holomorphic. It did not establish the converse. The distinction is useful at this stage of the course: a power series gives an explicit local representation and a direct way to compute derivatives, while holomorphicity is defined by the existence of complex derivatives. Do not treat these definitions as interchangeable unless the needed equivalence has been proved.

Check Your Understanding

Use the definitions, theorem, and examples in this tutorial to answer the following questions.

What does it mean for a function to be analytic at $z_0$?
Why does a power series define a holomorphic function at every point strictly inside its radius of convergence?
For $E(z)=\sum_{n=0}^{\infty}z^n/n!$, what is $E'(z)$, and why does the series converge for every complex $z$?
What is the radius of convergence of the displayed series for $1/(2-z)$?
How does the uniqueness theorem rule out two different coefficient sequences representing the same function near the same center?

What You'll Learn