Blog Entries: 5

## Taylor Series

$$\frac {1}{1-z} = \sum_{n=0}^{\infty} (\frac {(z-i)^n}{(1-i)^{n+1}}$$
we have to prove that, its our problem. So we start and we get,
$$\frac {1}{1-z} = \frac {1}{(1-i)-(z-i)} = \frac {1}{1-i} (\frac {1}{1- \frac {z-i}{1-i}})$$
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?

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 Recognitions: Homework Help Science Advisor You can use a Taylor series but it's also equivalent to a Binomial expansion which you should be able to handle. Here's a third way: $$\frac {1}{1-x} = \frac {1-x+x}{1-x} = 1 + \frac {x}{1-x}$$ and continue to replace the $\frac {1}{1-x}$ with $1 + \frac {x}{1-x}$ on the right side to obtain the desired expansion. Notice that -1 < x < 1 is required for convergence.
 Blog Entries: 5 does it matter then that z is a complex variable? it doesn't seem to make sense to say -1< z <1 if z is complex right? so then? and anyway, it says to use taylor series...

Recognitions:
Homework Help

## Taylor Series

$$\frac {1}{1-z} = \frac {1}{1-i} (\frac {1}{1- \frac {z-i}{1-i}})$$

Apply geometric series expansion to the right hand side. QED.

Recognitions:
Homework Help
 Blog Entries: 5 Alright, so i guess, $$\frac {1}{1-i} (\frac {1}{1- \frac {z-i}{1-i}})= \frac {1}{1-i} (1 + \frac {z-i}{1-i} +({ \frac {z-i}{1-i})^2 + .... + (\frac{z-i}{1-i})^n ) = \frac {1}{1-i} \sum_{n=0}^{\infty}(\frac {1}{1- \frac {z-i}{1-i}})^n= \sum_{n=0}^{\infty} (\frac {(z-i)^n}{(1-i)^{n+1}}= \frac {1}{1-z}$$ look about right? i've never done series much before, so i was just a bit lost, but i think that seems right if i use the power series as mentioned. only question: are there restrictions on z? i'm not really sure how to find them if so.
 Recognitions: Homework Help Science Advisor Gale, That is correct. It would simplify matters to replace $\frac {z-i}{1-i}$ with something like $w$ (call it whatever you want) so that you are expanding $\frac {1}{1-w}$ in your original equation. Then replace $w$ with $\frac {z-i}{1-i}$ when you're done. And, yes, there is a restriction on z which amounts to the absolute value of $w$ being less than 1.