Finding power series

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

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

Solution:

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 singularities are at z =1 and z = 2.

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

Solving for A and B (using standard algebraic methods or the cover-up method): $f(z) = \frac{1}{z-2} - \frac{1}{z-1} = \frac{1}{1-z} - \frac{1}{2-z}$

We are going to use the standard Geometric Series formula: $\frac{1}{1-w} = \sum_{n=0}^{\infty} w^n, \quad \text{valid for } |w| < 1$.

$\frac{1}{1-z} - \frac{1}{2-z} =[\text{Factor out 2: }] \frac{1}{1-z}-\frac{1}{2}(\frac{1}{1-\frac{z}{2}})$

$\frac{1}{2-z} = \frac{1}{2(1 - \frac{z}{2})} = \frac{1}{2} \cdot \frac{1}{1 - \frac{z}{2}}$

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 parts of the sum must converge simultaneously, 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).

$f(z) =\sum_{n=0}^{\infin} z^n(1 -\frac{1}{2}(\frac{1}{2^n})) =\sum_{n=0}^{\infin} (1 - \frac{1}{2^{n+1}})z^n$. Radius of Convergence: R = 1 (valid for $|z| < 1$).

Proposition. If f and g are analytic functions on a region $\Omega \subseteq \mathbb{C}$ and $fg \equiv 0, \forall z \in \Omega$, then either $f \equiv 0, \forall z \in \Omega$ or $g \equiv 0, \forall z \in \Omega$.

Analytic functions are “rigid.” Unlike smooth real functions (which can be zero in one place and non-zero immediately next to it), if an analytic function is zero on a small patch (or even a sequence of points converging to a limit), it is forced to be zero everywhere in that connected region.

Proof.

Assume $f \not\equiv 0 \text{ on } \Omega$, we must prove that this implies $g \equiv 0 \text{ on } \Omega$.

Since f is analytic and not identically zero on the connected region $\Omega$, its zeros are isolated. 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 $f(z) \neq 0 \forall z, 0<|z-z_0| < r$ (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 $f(z) \ne 0, \forall z \in B'(z_0, r)$.

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 on a set with a limit point $z_0$, so by the identify theorem we know that $g \equiv 0 \text{ on } B'(z_0, r)$. 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.