Week 7: Complex Power Series

This week, we will talk a lot about power series over $\mathbb{R}$ and $\mathbb{C}.$ The Weierstrass M-Test, which was proven last time, is what will help us take advantage of power series expansions of analytic functions. To understand the full power of power series, it is extremely useful to deal with complex quantities, so we begin with a brief review of complex numbers.

Definition:

For $i = \sqrt{-1},$ define the complex numbers $\mathbb{C} = \{a+bi: a,b \in \mathbb{R}\},$ with elementary operations given by

$(a+bi) \pm (c+di) = (a+c) \pm (b+d)i, (a+bi)(c+di) = (ac-bd)+(ad+bc)i,$

and

$\frac{a+bi}{c+di} = \frac{a+bi}{c+di} \cdot \frac{c-di}{c-di} = \frac{(ac+bd) + (bc-ad)i}{c^2+d^2}.$

For a complex number $z=a+bi,$ define the real and complex parts $\text{Re}(z) = a \in \mathbb{R}, \text{Im}(z) = b \in \mathbb{R},$ i.e. $z = \text{Re}(z) + i \text{Im}(z).$ Moreover, define the conjugate $\overline{z} = a-bi$ and the absolute value $|z| = \sqrt{a^2+b^2}$ (which defines a norm and therefore a metric on $\mathbb{C}$ ). Note the identity $z \overline{z} = |z|^2.$

Remark:

Note that the complex numbers are equivalent (isomorphic) as a metric space to the list of tuples $(\text{Re}(z),\text{Im}(z)) \in \mathbb{R}^2$ with the $l^2$ metric. Thus, $(\mathbb{C},d) \cong (\mathbb{R}^2,l^2)$ as metric spaces.
Recall that complex numbers can be represented in polar form $z = a+bi = r \cos \theta + i r \sin \theta = r(\cos \theta + i \sin \theta),$ where $r = |z|$ and $\theta = \arctan\left(\frac{b}{a}\right).$ By using the multiplication rules for complex numbers and trig identities for the sine and cosine of a sum of angles, one recognizes that

$r_1 (\cos \theta_1 + i \sin \theta_1) (\cos \theta_2 + i \sin \theta_2) = r_1 r_2 (\cos (\theta_1 + \theta_2) + i \sin(\theta_1 + \theta_2)).$

Motivated by this, we introduce Euler’s notation for complex numbers

$z = r(\cos \theta+i \sin \theta) =re^{i \theta}.$

Then, one easily checks that $\overline{z} = r(\cos \theta - i \sin \theta) = re^{-i\theta}.$ Similarly, to real functions $f:\mathbb{R} \to \mathbb{R},$ one can define complex-valued functions $f:\mathbb{C} \to \mathbb{C}$ taking values in the complex numbers. One can thus also take derivatives of continuous complex functions by treating $z$ as a variable, e.g. $\frac{d}{dz} z^2 = 2z, \frac{d}{dz} \frac{1}{z} = -\frac{1}{z^2},$ etc.

We are now able to approach power series in full generality.

Definition:

A complex power series centered around $z=a$ is an expression of the form $\sum_{n=0}^\infty a_n (z-a)^n, a_n,a,z \in \mathbb{C},$ which can be operated on using

$\sum_{n=0}^\infty a_n (z-a)^n + \sum_{n=0}^\infty b_n (z-a)^n = \sum_{n=0}^\infty (a_n+b_n) (z-a)^n$

and

$\sum_{n=0}^\infty a_n (z-a)^n \times \sum_{n=0}^\infty b_n (z-a)^n = \sum_{n=0}^\infty \left(\sum_{k=0}^n a_k b_{n-k}\right) (z-a)^n.$

If $z$ is a real number, the power series is called a real power series.

Remark:

Convergence of complex power series relies heavily on one of the convergence tests covered in 131A, namely, the Absolute Convergence Test, which states that if $\sum_{n=0}^\infty |a_n z^n|$ converges, then $\sum_{n=0}^\infty a_n z^n$ converges.

Proposition:

Suppose $f(z) = \sum_{n=0}^\infty a_n (z-a)^n$ is a complex power series. Then, if the power series converges for at least one $z$ such that $|z-a| = r,$ then it converges absolutely for $|z-a|<r.$

Proof:

Suppose $|z_0-a|<r.$ Then, since $\sum_{n=0}^\infty a_n(z-a)^n$ converges, $a_n(z-a)^n \to 0,$ so in particular, the terms of the sum are bounded. Suppose $|a_n(z-a)^n| \leq M$ for some $M>0.$ Then, $k = \frac{|z_0-a|}{|z-a|}<1,$ so

$|a_n(z_0-a)^n| = |a_n| (k|z-a|)^n\leq Mk^n,$

which gives a convergent geometric series since $k<1.$ Thus, by the Absolute Convergence Test, the series converges absolutely for $|z-a|<r.$

Definition:

Define the radius of convergence of a power series $R = \sup_{r \in \mathbb{R}}\{|z-a|: \sum_{n=0}^\infty a_n |z-a|^n \text{ converges }\}.$ Then, the power series converges for $|z-a|<r.$

Example:

The familiar geometric series $f(z) = \sum_{n=0}^\infty z^n$ converges for $|z|<1,$ since $\sum_{n=0}^\infty r^n = \frac{1}{1-r}$ for $|r|<1,$ showing that $f(z) = \frac{1}{1-z}$ on the open unit ball $B(0,1)$ in the complex plane.

Remark:

Note that the radius of convergence $R$ defines an open ball $B(a,R)$ on which the series converges. By definition, the series diverges for $|z-a|>R,$ but on the boundary $|z|=R,$ one has to use other techniques to determine whether the power series converges or diverges. For instance, in the previous example, for $z=1, \sum_{n=0}^\infty 1^n = \infty$ diverges.

Definition:

For a sequence $(x_n)_{n=1}^\infty \subset \mathbb{R}$ of real numbers, we define the limit superior and limit inferior to be

$\limsup_{n \to \infty} a_n = \lim_{n \to \infty} \sup_{k \geq n} a_n$

and

$\liminf_{n \to \infty} a_n = \lim_{n \to \infty} \inf_{k \geq n} a_n.$

Proposition:

If $(x_n)_{n=1}^\infty$ is a sequence of real numbers, $\lim_{n \to \infty} x_n$ exists if and only if $\limsup_{n \to \infty} x_n = \liminf_{n \to \infty} x_n,$ in which case

$\limsup_{n \to \infty} x_n = \liminf_{n \to \infty} x_n = \lim_{n \to \infty} x_n.$

Proof:

Exercise.

Proposition:

The radius of convergence of a power series $\sum_{n=0}^\infty a_n (z-a)^n$ is given by $\frac{1}{R} = \limsup_{n \to \infty} |a_n|^{\frac{1}{n}}.$

Proof:

We compare the series to a geometric series, which we know converges for $|z|<1.$ We thus want

$\sum_{n=0}^\infty |a_n z_n| = \sum_{n=0}^\infty (|a_n|^{\frac{1}{n}}|z|)^n \leq \sum_{n=0}^\infty r^n<\infty,$

which converges for $|a_n|^{\frac{1}{n}}|z|<r<1,$ or equivalently, for

$\frac{1}{|z|}>\frac{1}{|a_n|^{\frac{1}{n}}}.$

Since this has to be true as $n \to \infty,$ one has to take the $\lim \sup$ on both sides. Set $R= \frac{1}{\limsup_{n \to \infty}|a_n|^{\frac{1}{n}}}.$ One notes that for $|z|>R,$ comparison with a geometric series yields

$\infty = \sum_{n=0}^\infty |a_n (R+\epsilon)^n|\leq \sum_{n=0}^\infty |a_n z^n|,$

so the series diverges for $|z|>R$ and converges for $|z|<R.$ Thus, $R$ is the radius of convergence.
Another convergence test often comes in handy.

Proposition (Ratio Test):

The radius of convergence of $\sum_{n=1}^\infty a_n (z-a)^n$ is $\lim_{n \to \infty} \left|\frac{a_n}{a_{n+1}}\right|$ if the value of the limit exists.

Proof:

Exercise.

One notes that in the example above we were able to define a continuous function using a power series. One question to ask is when is the function defined by a power series continuous. The answer comes by using the Weierstrass M-Test.

Proposition:

Suppose $f(x) = \sum_{n=0}^\infty a_n (z-a)^n$ is a power series with radius of convergence $R.$ Pick $0<r<R.$ Then, if $\sum_{n=0}^\infty a_n r^n$ converges, the power series converges uniformly to a continuous function $f$ on $\overline{B(a,r)}.$

Proof:

Clearly, $a_n z^n$ is a continuous function with $\|a_n (z-a)^n\|_{l^\infty} \leq a_n r^n.$ Thus, an application of the Weierstrass M-Test immediately yields the result.

Remark:

This proposition implies that on any closed ball contained inside $B(0,R),$ where $R$ is the radius of convergence of the power series, a power series defines a continuous function $f.$