Taylor's Theorem for Holomorphic Functions

Every holomorphic function admits a convergent Taylor series expansion in any disk contained in its domain of definition. This fundamental result follows from the Cauchy integral formula.

Statement

Theorem5.9Taylor's theorem

Let $f$ be holomorphic on a domain $D$ and let $z_0 \in D$ . Then $f$ has a power series expansion

$f(z) = \sum_{n=0}^{\infty} a_n(z - z_0)^n, \qquad a_n = \frac{f^{(n)}(z_0)}{n!}$

converging absolutely and uniformly on compact subsets of the largest open disk $D(z_0, R) \subset D$ , where $R = \text{dist}(z_0, \partial D)$ .

Proof

Let $r < R$ and $\gamma = \{|w-z_0| = r\}$ (positively oriented). For $|z-z_0| < r$ , the Cauchy integral formula gives:

$f(z) = \frac{1}{2\pi i}\oint_\gamma \frac{f(w)}{w-z}\,dw.$

The key step is expanding the Cauchy kernel in a geometric series:

$\frac{1}{w-z} = \frac{1}{(w-z_0)-(z-z_0)} = \frac{1}{w-z_0}\cdot\frac{1}{1-\frac{z-z_0}{w-z_0}} = \sum_{n=0}^{\infty}\frac{(z-z_0)^n}{(w-z_0)^{n+1}}$

which converges uniformly in $w$ on $\gamma$ since $|z-z_0|/|w-z_0| = |z-z_0|/r < 1$ .

Substituting into the Cauchy formula and interchanging sum and integral (justified by uniform convergence):

$f(z) = \sum_{n=0}^{\infty} \left(\frac{1}{2\pi i}\oint_\gamma \frac{f(w)}{(w-z_0)^{n+1}}\,dw\right)(z-z_0)^n = \sum_{n=0}^{\infty}\frac{f^{(n)}(z_0)}{n!}(z-z_0)^n.$

Since this holds for every $r < R$ , the series converges on $D(z_0, R)$ .

Absolute convergence follows from the estimate $|a_n| \leq M_r/r^n$ where $M_r = \max_{|w-z_0|=r}|f(w)|$ , giving $|a_n(z-z_0)^n| \leq M_r(|z-z_0|/r)^n$ , a convergent geometric series for $|z-z_0| < r$ . $\blacksquare$

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Important Consequences

RemarkAnalytic equals holomorphic

Taylor's theorem proves the equivalence of analyticity and holomorphicity for complex functions. This is in sharp contrast to real analysis, where $C^\infty$ functions can fail to equal their Taylor series (e.g., $f(x) = e^{-1/x^2}$ at $x = 0$ ). In the complex setting, differentiability at each point forces convergence of the Taylor series.

ExampleStandard Taylor expansions

The following Taylor series converge on all of $\mathbb{C}$ (radius $R = \infty$ ):

$e^z = \sum_{n=0}^\infty \frac{z^n}{n!}, \qquad \sin z = \sum_{n=0}^\infty \frac{(-1)^n z^{2n+1}}{(2n+1)!}, \qquad \cos z = \sum_{n=0}^\infty \frac{(-1)^n z^{2n}}{(2n)!}.$

The series $\frac{1}{1-z} = \sum_{n=0}^\infty z^n$ has $R = 1$ , determined by the singularity at $z = 1$ .

ExampleRadius equals distance to nearest singularity

For $f(z) = \frac{1}{1+z^2}$ , the Taylor series centered at $z_0 = 0$ has $R = 1$ because the nearest singularities are at $z = \pm i$ with $|z_0 - (\pm i)| = 1$ . Centered at $z_0 = 2$ , the radius would be $R = |2 - i| = \sqrt{5}$ .

RemarkTerm-by-term operations

Within the disk of convergence, one may freely:

Differentiate term by term (preserving radius of convergence)
Integrate term by term
Multiply two power series using the Cauchy product
Compose power series (with appropriate radius conditions)