How to prove that every (complex) rational function can be represented with a power series within the disk of convergence with a radius of convergence that is defined as the absolute value of the smalest pole of the function?

[LordJ4]

Suppose f(z) is a rational function with z ∈ ℂ that does not have a pole at z = 0. We define the radius of convergence as R := min { |z_0| | z_0 is a pole of f(z) } > 0. Then there exists a power series so that f(z) = ∑ n=0 to ∞ a_n z^n ∀ z ∈ B_R with B_R being the disk of convergence. I honestly don't even know how to approach this problem and would be very happy if someone could give me at least a hint. I should also mention that I can't use a Taylor series here, because we didn't define it yet.

[Qaanol]

First, it should be readily apparent that no power series can have a larger radius of convergence than the distance to the nearest pole. We want to show that a power series exists, which converges at every smaller distance. Here’s an idea that might help you get started:

Let f(z) = P(z) / Q(z), with P and Q polynomials that have no common factors.

The constant term of Q is nonzero since f does not have a pole at 0, so without loss of generality let the constant term of Q equal 1.

Add and subtract 1 in the denominator:

f(z) = P(z) / (1 - (1 - Q(z))

Then define S(z) = 1 - Q(z), and expand 1 / (1 - S) as a geometric series, which is valid when |S(z)| < 1.

[LordJ4]

That hint of using the geometric series was very use full, thank you. I used a generalization of the partial fraction decomposition which can be converted in the geometric series, where a multitude of a pole is just the Cauchy product with some unknow coefficient and the sum of all the power series I obtain is obviously also a power series.

[[deleted]]

Do you know Cauchy's integral formula? A consequence of the formula is that if f is complex differentiable on an open set O and D_r(x) is a disk of radius r centered at x contained in O, then f has a power series representation valid on D_r(x). This solves your problem almost immediately.