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


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



  • $\begingroup$ 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

Let $z=1/u$ to get


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:



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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.