# Thread: Complex Analysis - Partial Fraction Decomposition

1. ## Complex Analysis - Partial Fraction Decomposition

Hey guys. This might not be quite the right place on the forum, but here goes. I have this assignment that I have almost completed, but I can't quite figure out the last bit. Could anyone give me a helping hand?

Prove that$\displaystyle F_0 = 0$, $\displaystyle F_1 = 1$ and $\displaystyle F_n =F_{n-1} + F_{n-2}$ (This is the sequence of fibonacci numbers: 0, 1, 1, 2, 3, 5, 8,...).

Find the partial fraction decomposition of the rational function and use it to prove Binet's formula:

(1)
$\displaystyle F_n = \frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^n - \left(\frac{1-\sqrt{5}}{2}\right)^n\right]$

Alright, I have shown that $\displaystyle \alpha_1 = \frac{-1 + \sqrt{5}}{2}$ and $\displaystyle \alpha_2 = \frac{-1 - \sqrt{5}}{2}$ are simple poles of $\displaystyle f(x)$ and I've found out that the partial fraction decomposition of $\displaystyle f(z)$ is:

(2)
$\displaystyle f(z) = \frac{\frac{1+\sqrt{5}}{2\sqrt{5}}}{z-\alpha_1} - \frac{\frac{1-\sqrt{5}}{2\sqrt{5}}}{z-\alpha_2}$

I just have no idea how to use that, to show (2) from (1). Can someone give me a clue?

/Morten

2. ## Re: Complex Analysis - Partial Fraction Decomposition

Find the partial fractions decomposition of what rational function? I don't see anywhere that you say what rational function you are talking about.

3. ## Re: Complex Analysis - Partial Fraction Decomposition

Arh sorry.. Apparently it only pasted half of the assignment into the post. Here is the first half of the assignment:

Prove that the function $\displaystyle z/(1-z-z^2)$ has simple poles at $\displaystyle z = (-1 \pm \sqrt{5})/2$ and determine the radius of convergence of the power series:

$\displaystyle \frac{z}{1-z-z^2} = \sum_{n = 0}^{\infty}F_nz^n$

So the rational function is $\displaystyle f(z) = \frac{z}{1-z-z^2}$

4. ## Re: Complex Analysis - Partial Fraction Decomposition

Surely you understand that the poles are where the denominator is 0...