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Given the geometric series

$\frac{1}{1-z} = \sum_{n=0}^n = 1 + z + z^2 + ...$

If there is a function $f(z)=\frac{1}{z+j}$ how would you get it's Taylor series about center z = 1? I have tried the following to get it into the form of $\frac{1}{1-(z-z_0)}$

$$f(z) = \frac{1}{1-(z-1)+j-2)}$$

but that is clearly not right, or i don't think it is anyway.

Could I get some hints as to how to proceed?

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    $$\dfrac{1}{z+j}=\dfrac{1}{1+j}\dfrac{1}{\frac{z-1}{j+1}+1}$$2012-06-21
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    @J.M. Thanks, but I still don't seem to get how this can be fitted into the Geometric Series, can you elaborate a little please?2012-06-21
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    You know the series expansion for $\dfrac1{1+w}$, yes? After doing this expansion, let $w=\dfrac{z-1}{j+1}$...2012-06-21
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    @J.M. Sorry but i don't get it. The last few questions I have been doing is more along the lines of $\frac{1}{3-2z}$ with center at 1. which rearranges to $$\frac{1}{1-2(z-1)}$$ which looks very similar to the geometric series $$\frac{1}{1-z}$$. I substitute 2(z-1) into z from geometric series and get my answer.2012-06-21
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    Still the same strategy. If you can expand $\dfrac1{1-w}$, you can also expand $\dfrac1{1+w}$...2012-06-21
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    @J.M. I don't actually know how to expand it, it's just given to me and I memorized it.2012-06-21
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    [Well then...](http://books.google.com/books?hl=en&id=OisInC1zvEMC&pg=PA368)2012-06-21

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