Proving $\frac{1}{1-z} = \sum_{n=0}^{\infty} (\frac {(z-i)^n}{(1-i)^{n+1}}$

[Gale]

$$\frac {1}{1-z} = \sum_{n=0}^{\infty} (\frac {(z-i)^n}{(1-i)^{n+1}}$$
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?

 

[Tide]

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.

 

[Gale]

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...

 

[benorin]

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

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

 

[shmoe]

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?

Then you'd talk about bounds on |z|

and anyway, it says to use taylor series...

have you tried directly finding the power series of 1/(1-z)? What point do you want to expand it about?

 

[Gale]

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.

 

[Tide]

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.