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$$f(z)=\frac{2z}{z^2+1}=\frac1{z-i} +\frac1{z+i}$$ Find Laurent series in powers of $z$ in the domain $|z|<1$.

So I got to find two Taylor series of the two terms in the function but how do you do that since there is $i$ instead of $1$. I did try to divide by $i$ but then it gives a series for a different domain.

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    $\begingroup$ Avoid saying asap, step by step, etc. in your question. And multiply your fraction by $\frac{i}{i}$. $\endgroup$ – Aaron Maroja May 25 '15 at 23:06
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For $|z|<|i|=1$ we have \begin{align*} \frac{1}{z-i}&=-\frac{1}{i-z}\\ &=-\frac{1}{i}\frac{1}{1-\frac{z}{i}}\\ &=-\frac{1}{i}\sum\limits_{k=0}^\infty\left(\frac{z}{i}\right)^k \end{align*} This is just using the geometric series. Where $\sum\limits_{k=0}^\infty\left(\frac{z}{i}\right)^k$ converges if $\left|\frac{z}{i}\right|<1$. (Note that this is equivalent to saying $|z|<1$ because $|i|=1$.)

You can do the same for $\frac{1}{z+i}$

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I think you have the right idea. Note that $|z/i|<1 \iff |z|<1$.

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  • $\begingroup$ Seriously? $|z|<i =1$???? $\endgroup$ – snowman May 25 '15 at 23:02
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    $\begingroup$ I'm saying that $|z/i|=|z|$. So the domain $|z/i|<1$ is the same as the domain $|z|<1$. $\endgroup$ – TorsionSquid May 25 '15 at 23:03

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