- #1
jostpuur
- 2,116
- 19
I have figured out a nice way to prove that if the complex numbers [itex]z_1,z_2,\ldots, z_N\in\mathbb{C}[/itex] are all distinct, then the equation
[tex]
\prod_{n=1}^N \frac{1}{z - z_n} = \sum_{n=1}^N \frac{\alpha_n}{z-z_n}
[/tex]
is true for all [itex]z\in\mathbb{C}\setminus\{z_1,z_2,\ldots, z_N\}[/itex], where the alpha coefficients have been defined by the formula
[tex]
\alpha_n = \underset{n'\neq n}{\prod_{n'=1}^N} \frac{1}{z_n - z_{n'}}
[/tex]
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.
[tex]
\prod_{n=1}^N \frac{1}{z - z_n} = \sum_{n=1}^N \frac{\alpha_n}{z-z_n}
[/tex]
is true for all [itex]z\in\mathbb{C}\setminus\{z_1,z_2,\ldots, z_N\}[/itex], where the alpha coefficients have been defined by the formula
[tex]
\alpha_n = \underset{n'\neq n}{\prod_{n'=1}^N} \frac{1}{z_n - z_{n'}}
[/tex]
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.