# Abstract Nonsense

## Complex Power Series

Point of Post: In this post we discuss the basic ideas behind when a complex power series converges and discuss the holomorphicity of functions representable by power series.

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Motivation

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We have now discussed the notion of holomorphic functions but beside’s some extremely trivial ones (like polynomials) we don’t have any good class of examples of such functions. In this post we describe what is the most informative example of holomorphic functions–power series. Why are power series the most informative example? Well we shall eventually see that the holomorphic functions are precisely those that are locally just power series.

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I assume anyone reading this is fairly well-acquainted with power-series and the related notions (the Weierstrass M-test, etc.) , for I will gloss over some stuff.

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Power Series

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Let $\{c_n\}$ be a sequence of complex numbers. We say that the series $\displaystyle \sum c_n$ converges if the sequence $\displaystyle \left\{\sum_{n=1}^{N}c_n\right\}$ converges in the usual sense. Since $\mathbb{C}\approx\mathbb{R}^2$ is complete we know that a sequence converges if and only if it’s Cauchy, and thus we see that $\displaystyle \sum_n c_n$ converges if and only if for every $\varepsilon>0$ there exists $N\in\mathbb{N}$ such that $k>N$ implies that $\displaystyle \left|\sum_{n=k}^{\infty}c_n\right|<\varepsilon$.

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Recall the notation that if $f:X\to\mathbb{C}$ with $X\subseteq\mathbb{C}$ then $\|f\|_\infty$, the infinity norm, is defined to be equal to $\displaystyle \sup_{x\in X}|f(x)|$. Clearly $\|f\|_\infty$ needn’t be finite.  Recall then that we say that a sequence of functions $\{f_n\}$ mapping $X\to\mathbb{C}$ converges pointwise to a function $f:X\to\mathbb{C}$ if for each $x_0\in X$ one has that $\lim f_n(x_0)=f(x_0)$. We say that $\{f_n\}$ converges uniformly to $f$ if for each $\varepsilon>0$ there exists some $N\in\mathbb{N}$ such that $n>N$ implies that $\|f_n-f\|<\varepsilon$. This says roughly that $f_n(x)\to f(x)$ “the same” for each $x\in X$. We say that $\{f_n\}$ is uniformly Cauchy if for all sufficiently large $m,n$ one has that $\|f_n-f_n\|<\infty$. It’s a fundamental fact that since $\mathbb{C}$ is complete that $\{f_n\}$ converges uniformly (to some function) if and only if it is uniformly Cauchy. It is a common fact that if $\{f_n\}$ is a sequence of uniformly Cauchy continuous functions $X\to\mathbb{C}$ then $f_n$ converges uniformly to some continuous function $X\to\mathbb{C}$.

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If $X$ is compact then $\|\cdot\|_\infty$ defines a norm on the space $C(X;\mathbb{C})$ of all complex valued continuous functions on $X$.

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When we speak of the (uniform) convergence of a power series $\displaystyle \sum_n a_n(z-z_0)^n$ we really are discussing the (uniform) convergence of the partial sum polynomials $\displaystyle S_N(z)=\sum_{n=1}^{N}a_n(z-z_0)^n$.

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Suppose we are given a power series $\displaystyle \sum_n a_n(z-z_0)^n$. We define the radius of convergence $R$ of the series to be

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$\displaystyle R=\frac{1}{\overline{\lim}\sqrt[n]{|c_n|}}$

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(here $\overline{\lim}$ means the limit supremum) where $R=0$ if $\overline{\lim}\sqrt[n]{|c_n|}=\infty$ and $R=\infty$ if $\overline{\lim}\sqrt[n]{|c_n|}=0$.

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Theorem: Let $\displaystyle \sum_n a_n(z-z_0)^n$ be a series with radius of convergence $R$. Then, $\displaystyle \sum_n a_n(z-z_0)^n$ and $\displaystyle \sum_n|a_n(z-z_0)^n|$ converge uniformly on $\overline{B_r(z_0)}$ (here $B_r(z_0)$ denotes the ball of radius $r$ centered at $z_0$ and the overline means closure) for each $0\leqslant r. Moreover, $\displaystyle \sum_n a_n(z-z_0)^n$ diverges for $z\notin B_R(z_0)$.

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Note that nothing is said about points on $\partial B_R(z_0)$ (here $\partial$ means boundary)–this is because nothing really can be said. It’s totally possible that $\displaystyle \sum_n a_n(z-z_0)^n$ converges nowhere, everywhere, or just some places on $\partial B_R(z_0)$.

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Now onto our first indication that power series are important for the discussion of holmorphic functions. Let $\Omega\subseteq\mathbb{C}$. Say that $f:\Omega\to\mathbb{C}$ is locally representable by power series if for every $z_0\in\Omega$ there exists a ball $B_r(z_0)\subseteq\Omega$ and a power series $\displaystyle \sum a_n(z-z_0)^n$ such that $\displaystyle f(z)=\sum a_n(z-z_0)^n$ on $B_r(z_0)$. The big theorem then is the following:

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Theorem: Let $\Omega\subseteq\mathbb{C}$ be open. Then, if $f$ is locally representable by power series one has that $f\in\mathcal{O}(\Omega)$, and moreover if $f$ is locally $\displaystyle \sum_n a_n(z-z_0)^n$ at $z_0$ then $f'$ is locally $\displaystyle \sum_n n a_n (z-z_0)^n$ at $z_0$.

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I’ll leave you to look up the theorem (it’s in [3] for example)–it’s just an $\varepsilon$-chase. The important thing to note though is that since $f'$ is locally representable by power series it is, again, holmorphic! Thus, we actually find the following:

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Theorem: If $f$ is locally representable by power series then $f$ is infinitely differentiable on $\Omega$, and if $f$ is locally $\displaystyle \sum a_n (z-z_0)^n$ then $f^{(k)}$ is locally $\displaystyle \sum n(n-1)\cdots(n-(k+1))(z-z_0)^{n-k}$. In particular, $f^{(k)}(z_0)=k!a_k$.

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In particular, if two series are equal, say $\displaystyle \sum a_n(z-z_0)^n=f(z)=\sum b_n(z-z_0)^n$ on some open set $\Omega$ then we know that $a_k=\frac{1}{k!}f^{(k)}(a)=b_k$. In particular:

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Theorem: Let $\Omega\subseteq\mathbb{C}$ be open. If $\displaystyle \sum a_n(z-z_0)^n=\sum b_n (z-z_0)^n$ on $\Omega$ then $a_n=b_n$ for all \$latex n.

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References:

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[1] Greene, Robert Everist, and Steven George Krantz. Function Theory of One Complex Variable. Providence: American Mathematical Society, 2006. Print.

[2] Conway, John B. Functions of One Complex Variable I. New York: Springer-Verlag, 1978. Print.

[3] Rudin, W. Real and Complex Analysis. New York,NY: McGraw-Hill, 1988. Print.

[4]  Ahlfors, Lars V. Complex Analysis; an Introduction to the Theory of Analytic Functions of One Complex Variable. New York: McGraw-Hill, 1966. Print.