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How to find the Laurent series for |z|>1 for $$\dfrac{e^{1/z}}{z^2-1} ?$$ Firstly, how to decide if given function has laurent series expansion in specified domain or not? I just did the long division .but I am not sure if that is the expected answer or not. Please help

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$$|z|>1\implies\frac1{|z|}<1\implies\frac{e^{1/z}}{z^2-1}=\frac1{z^2}\cdot e^{1/z}\cdot\frac1{1-\frac1{z^2}}=$$

$$=\frac1{z^2}\left(1+\frac1z+\frac1{2!z^2}+\frac1{3!z^3}+\ldots\right)\left(1+\frac1{z^2}+\frac1{z^4}+\ldots\right)=$$

$$\frac1{z^2}+\frac1{z^3}+\frac3{2z^4}+\frac4{3z^5}+\ldots$$

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  • $\begingroup$ Doing just long division will not account that domain is |z|>1,right?we have to expand the laurent series in the annular region: infinity>|z|>1 so,this domain must be taken into account.am I right? $\endgroup$ – Awani Khodkumbhe Apr 30 '17 at 12:19
  • $\begingroup$ @AwaniKhodkumbhe I'm not sure I understand your question: what "long division" are you talking about? I just took the domain $\;|z|>1\;$ given by you and developed both the exponential function (which has a power series that converges everywhere) and that part of the geometric series. $\endgroup$ – DonAntonio Apr 30 '17 at 12:32
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Let $z=1/u$ to get

$$\frac{e^uu^2}{1-u^2}$$

since we want $|z|>1$, then we want $|u|<1$, that is, the Laurent series at $u=0$. This is very easy, just notice that:

$$e^u=1+u+\frac12u^2+\frac16u^3+\dots$$

$$\frac1{1-u^2}=1+u^2+u^4+u^6+\dots$$

and multiply it all together, then let $u=1/z$ to turn it back in terms of $z$.

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