0
$\begingroup$

I am having some confusion about Laurent series at infinity and essential singularities. Consider the function: $$f(z)=\frac{1}{1-z}$$ which has a Laurent Series about infinity of: $$f(z)=-\frac{1}{z}-\frac{1}{z^2}-...$$ From this it would appear that $f(z)$ would have an essential singularity. However, $$f(t)=\frac{t}{t-1}$$ Where $t=1/z$ does not have an essential singularity at $t=0$ which would mean that $f(z)$ does not have an essential singularity at $z=\infty$. Which of my reasonings is right and why is the other wrong?

$\endgroup$
  • 1
    $\begingroup$ You mix up $t=0$ (i.e. $z=\infty$) and $z=0$. They are completely different points. Your Laurent series converges for $|z|>1$ where $z=0$ does not belong to, so the expansion is not valid around $z=0$. $\endgroup$ – A.Γ. Aug 24 '16 at 6:44
  • $\begingroup$ @A.G. so $f(z)=-\frac{1}{z}-\frac{1}{z^2}-...$ is not the Laurent series about infinity? $\endgroup$ – Quantum spaghettification Aug 24 '16 at 6:54
  • $\begingroup$ It is. But what "around infinity" has to do with "near the zero"? $\endgroup$ – A.Γ. Aug 24 '16 at 6:57
  • $\begingroup$ @A.G. Sorry I don't quite know why you are bringing zero into this? If $f(z)$ is the Laurent series about $\infty$ simply by the virtue that it has a non-terminating principal part, means it is a singularity? $\endgroup$ – Quantum spaghettification Aug 24 '16 at 15:26
  • $\begingroup$ When you say that $-\frac{1}{z}-\frac{1}{z^2}-...$ has a principal part you mean implicitly that it is a Laurent series at zero (note that at infinity it is analytical, so no principal part). This is how $z=0$ comes to play. In general, "principal part = negative powers" works for finite numbers only. For infinity the correct way to define it is the substitution $t=1/z$ and reducing to the case of $t=0$. $\endgroup$ – A.Γ. Aug 25 '16 at 7:06

Your Answer

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

Browse other questions tagged or ask your own question.