- #1
- 676
- 2
[tex] \frac {1}{1-z} = \sum_{n=0}^{\infty} (\frac {(z-i)^n}{(1-i)^{n+1}}[/tex]
we have to prove that, its our problem. So we start and we get,
[tex] \frac {1}{1-z} = \frac {1}{(1-i)-(z-i)} = \frac {1}{1-i} (\frac {1}{1- \frac {z-i}{1-i}})[/tex]
and then i think we're supposed to go to a power series or something, but i don't know, I'm not sure how to get to there, from here. help?
we have to prove that, its our problem. So we start and we get,
[tex] \frac {1}{1-z} = \frac {1}{(1-i)-(z-i)} = \frac {1}{1-i} (\frac {1}{1- \frac {z-i}{1-i}})[/tex]
and then i think we're supposed to go to a power series or something, but i don't know, I'm not sure how to get to there, from here. help?