# I Partial Fraction Challenge

1. Feb 17, 2017

### jostpuur

I have figured out a nice way to prove that if the complex numbers $z_1,z_2,\ldots, z_N\in\mathbb{C}$ are all distinct, then the equation

$$\prod_{n=1}^N \frac{1}{z - z_n} = \sum_{n=1}^N \frac{\alpha_n}{z-z_n}$$

is true for all $z\in\mathbb{C}\setminus\{z_1,z_2,\ldots, z_N\}$, where the alpha coefficients have been defined by the formula

$$\alpha_n = \underset{n'\neq n}{\prod_{n'=1}^N} \frac{1}{z_n - z_{n'}}$$

I would like to leave the proof of this result as challenge to you guys, and I'm not in a need for advice myself at this point. Of course if somebody proves this in a way that is different from my proof, then I'm still reading.

2. Feb 19, 2017

### Staff: Mentor

[ SPOILER="spoiler" ]

There might be a simpler way, but you can prove it by induction. The base case is $N=1$. Then we need to show that

$\Pi_{n=1}^1 \frac{1}{z-z_n} = \sum_{n=1}^1 \frac{\alpha_n}{z-z_n}$

That's pretty obvious. Now, suppose it is true for $N$. Then we have:

$\Pi_{n=1}^N \frac{1}{z-z_n} = \sum_{n=1}^N \frac{\alpha_n}{z-z_n}$

Now, multiply both sides by $\frac{1}{z-z_{N+1}}$. On the left side, we have:

$\Pi_{n=1}^{N+1} \frac{1}{z-z_n}$

On the right side, we have:

$\sum_{n=1}^N \frac{\alpha_n}{z-z_n} \frac{1}{z-z_{N+1}}$

Now, we use the fact that:

$\frac{1}{z-z_n} - \frac{1}{z-z_{N+1}} = \frac{z_n -z_{N+1}}{(z-z_n) (z-z_{N+1})}$

So $\frac{1}{z_n - z_{N+1}} [\frac{1}{z-z_n} - \frac{1}{z-z_{N+1}}]= \frac{1}{(z-z_n) (z-z_{N+1})}$

So we can rewrite the right-hand side as

$\sum_{n=1}^N \frac{\alpha_n}{z_n - z_{N+1}} [\frac{1}{z-z_n} - \frac{1}{z-z_{N+1}}]$

$= [\sum_{n=1}^N \frac{\bar{\alpha_n}}{z - z_n}] + \frac{\bar{\alpha_{N+1}}}{z-z_{N-1}}$

where $\bar{\alpha_n} = \frac{\alpha_n}{z_n - z_{N+1}}$ and where $\bar{\alpha_{N+1}} = \sum_{n=1}^N -\frac{\alpha_n}{z_n - z_{N+1}}$

So we can write the right-hand side as $\sum_{n=1}^{N+1} \frac{\bar{\alpha_n}}{z - z_n}$

So the only thing left is to prove that $\bar{\alpha_n}$ has the right form, which is beyond what I have the energy for right now.
[ /SPOILER ]