Power Series and the Integrity of Analytic Functions

If you’re going through hell, keep on going, Winston Churchill

Power Series Expansion: A Worked Example

Problem. Find the power series expansion of the function $f(z) = \frac{1}{z^2-3z+2}$ centered at 0 and determine its radius of convergence.

Solution:

Step 1: Locate the Singularities

First, we factor the denominator to identify the singularities (poles) of the function: $\frac{1}{z^2-3z+2} = \frac{1}{(z-1)(z-2)}$.

The function has isolated singularities (simple poles) at $z = 1$ and $z = 2$.

In complex analysis, the radius of convergence $R$ for a Taylor series centered at $z_0$ is the distance to the nearest singularity.

Since we are expanding about $z_0 = 0$, the radius of convergence will be the distance from 0 to the nearest singularity, that is, 1. Here, $z_0 = 0$, so we expect $R = \text{dist}(0, 1) = 1$.

So we already expect the radius of convergence to be $\boxed{R = 1}.$

Step 2: Partial Fraction Decomposition

Next, we use Partial Fraction Decomposition to split the function into simpler terms (components) $\frac{1}{(z-1)(z-2)} = \frac{A}{z-1} + \frac{B}{z-2}$

Solving 1 = A(z - 2) + B(z - 1).

Setting $z = 2: 1 = A(0) + B(1) \Rightarrow B = 1.$

Setting $z = 1: 1 = A(-1) + B(0) \Rightarrow A = -1.$

Thus, $\frac{1}{(z-1)(z-2)} = \frac{A}{z-1} + \frac{B}{z-2} = \frac{1}{z-2} - \frac{1}{z-1} = \frac{1}{1-z} - \frac{1}{2-z}.$

Step 3: Series Expansion

We recall the standard geometric series: $\frac{1}{1-w} = \sum_{n=0}^{\infty} w^n$ for $|w| < 1$.

1. First term: $\frac{1}{1-z} = \sum_{n=0}^{\infty} z^n$, valid for $|z| < 1$.
2. Second term: To use the geometric formula, we must factor out a $2$ from the denominator: $$-\frac{1}{2-z} = -\frac{1}{2(1 - \frac{z}{2})} = -\frac{1}{2} \sum_{n=0}^{\infty} \left(\frac{z}{2}\right)^n = -\sum_{n=0}^{\infty} \frac{z^n}{2^{n+1}}$$ This is valid for $|\frac{z}{2}| < 1$, or $|z| < 2$.

Step 4: Combine and Find Radius

Next, we expand both terms using the geometric series formula:

$f(z)=\sum_{n=0}^{\infin}z^n -\frac{1}{2}(\sum_{n=0}^{\infin}(\frac{z}{2})^n)$ where these series converges for |z| < 1 and $|\frac{z}{2}| \lt 1 \implies |z| \lt 2$ respectively.

For the power series to exist, both individual series must converge, and the intersection of both regions is |z| < 1 (we could have reasoned alternatively that the radius of convergence is always the distance from the center z = 0 to the nearest singularity, in this particular case 1).

Result: $f(z) =\sum_{n=0}^{\infin} z^n(1 -\frac{1}{2}(\frac{1}{2^n})) =\sum_{n=0}^{\infin} (1 - \frac{1}{2^{n+1}})z^n$ with radius of convergence R = 1.

This principle is fundamental:

The radius of convergence of a Taylor series equals the distance to the nearest singularity.

The Integrity of Analytic Functions

A fascinating algebraic property of analytic functions is that they form an integral domain. That is, they have no “zero divisors.”

Proposition. Let $f$ and $g$ be analytic functions on a connected region $\Omega \subseteq \mathbb{C}$. If $f(z)g(z) = 0$ for all $z \in \Omega$, then either $f \equiv 0$ or $g \equiv 0$ on $\Omega$.

Intuition: 💡 Analytic functions are “rigid” and cannot vanish “sporadically.” Unlike smooth real functions $C^\infty$ (which can be zero in one interval and non-zero immediately on another), if an analytic function is zero on a small open set, it is forced to be zero everywhere in that connected region. In other words, an analytic function cannot “turn off” on an open set without being zero everywhere.

Proof.

Assume $f$ is not identically zero ($f \not\equiv 0$); we must show that $g \equiv 0$.

Step 1. Zeros of Analytic Functions Are Isolated

Since f is analytic and not identically zero on the connected region $\Omega$, its zeros are isolated. This means the set $Z_f = \{z \in \Omega : f(z) = 0\}$ contains no limit points within $\Omega$.

Let $z_0 \in \Omega$, then there exists r > 0 such that either:

1. By continuity, $f(z) \neq 0, \forall z, |z-z_0| < r$ (if $f(z_0)\neq 0, z_0$ is a zero), or
2. By the zeros are isolated theorem, $f(z_0) = 0$ and $f(z) \neq 0 \quad \forall z: 0 < |z - z_0| < r$ (if $z_0$ is a zero, which is isolated).

Therefore, there is an r > 0 such that $B(z_0, r) \subseteq \Omega$ ($\Omega$ is a region, hence open) and $f(z) \ne 0, \forall z \in B'(z_0, r)$.

In either case, there is a punctured neighborhood where $f(z) \neq 0.$

Step 2: Use the Product Condition

However, we are given the condition $fg \equiv 0 \text{ on } \Omega$. Restricting this to our disk, since the product is zero and the first factor is not zero, the second factor must be zero, $g(z) = 0, \forall z \in B'(z_0, r)$. g(z) is 0 in that punctured set.

Thus, g vanishes on a set with a limit point $z_0$.

Step 3: Apply the Identity Theorem

Thus, $g \equiv 0 \text{ on } B'(z_0, r)$. However, g vanishes not only on the punctured ball $B'(z_0, r)$, but on the entire (connected) region $\Omega$ that contains the ball.

The Identity Theorem states that if an analytic function on a region vanishes on a set with a limit point inside its domain, it must vanish on the entire connected domain or region where it is analytic.

Conclusion

Either $f \equiv 0 \quad \text{or} \quad g \equiv 0.$ $\boxed{\text{Analytic functions have no zero divisors.}}$

Conceptual Insight

This proposition shows:

• The ring of analytic functions on a connected region has no zero divisors.
• Analytic functions behave more like polynomials than like arbitrary smooth functions.
• The Identity Theorem is one of the deepest rigidity results in complex analysis. It is this rigidity that makes complex analysis so powerful.
• This property relies heavily on $\Omega$ being a region (an open connected set). If $\Omega$ consisted of two disjoint disks $D_1$ and $D_2$, we could define $f=0$ on $D_1, f=1$ on $D_2$ and $g=1$ on $D_1, g=0$ on $D_2$. The product would be zero everywhere, but neither function would be identically zero.