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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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$$\dfrac{1}{z+j}=\dfrac{1}{1+j}\dfrac{1}{\frac{z-1}{j+1}+1}$$ – J. M. Jun 21 '12 at 9:48
@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? – Synia Jun 21 '12 at 9:56
You know the series expansion for $\dfrac1{1+w}$, yes? After doing this expansion, let $w=\dfrac{z-1}{j+1}$... – J. M. Jun 21 '12 at 9:59
@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. – Synia Jun 21 '12 at 10:07
Still the same strategy. If you can expand $\dfrac1{1-w}$, you can also expand $\dfrac1{1+w}$... – J. M. Jun 21 '12 at 10:10

Take any $a\in\Bbb C$. For $j\neq -a$, we note that $$\begin{align}z+j &= z-a+j+a\\ &= (j+a)\left(\frac{z-a}{j+a}+1\right)\\ &= (j+a)\left(1-\left(-\frac{z-a}{j+a}\right)\right).\end{align}$$ Note that it is precisely in the disk $|z-a|<|j+a|$ that we have $$\left|-\frac{z-a}{j+a}\right|=\frac{|z-a|}{|j+a|}<1.$$ Since $$\frac{1}{1-w}=\sum_{k=0}^\infty w^k$$ for $|w|<1$, then letting $w=-\frac{z-a}{j+a}$, we have in the disk $|z-a|<|j+a|$ that

$$\begin{align} \frac{1}{z+j} &= \frac{1}{j+a}\frac{1}{1-w}\\ &= \frac{1}{j+a}\sum_{k=0}^\infty w^k\\ &= \frac{1}{j+a}\sum_{k=0}^\infty \frac{(-1)^k(z-a)^k}{(j+a)^k}\\ &= \sum_{k=0}^\infty\frac{(-1)^k}{(j+a)^{k+1}}(z-a)^k,\\ \end{align}$$

giving us the Taylor series for $\dfrac{1}{z+j}$ about $z=a$ (where $j\neq -a$).

The above general approach works just fine for $a=1$ (which is the particular case you're considering), so long as $j\neq -1$. Of course, if $j=-1$, then we'd be trying to find the Taylor series expansion of $\dfrac{1}{z-1}$ about $z=1$--an impossible task, since $\dfrac{1}{z-1}$ fails to be defined at $z=1$.

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