7
$\begingroup$

I must find the residues of $z^2\sin(\frac{1}{z})$ at $z = 0$.

Since $z = 0$ seems to be an Essential Singularity, i'm not sure how I can continue to find the residue of the function. Usually I am able to apply the Taylor Series and then find the $z^{-1}$ coefficient, but in this case I do not get a $z^{-1}$ term.

$\endgroup$
7
$\begingroup$

Since $w\mapsto\sin w$ is an entire function, we can write it as its Taylor series around $0$: $$ \sin w=\sum_{n=0}^{+\infty}\frac{(-1)^n}{(2n+1)!}w^{2n+1} $$ thus writing $w=\frac1z$ we get $$ \sin\frac1z=\sum_{n=0}^{+\infty}\frac{(-1)^n}{(2n+1)!}\frac1{z^{2n+1}} $$ which is valid on the punctured plane $\Bbb C^{\times}$, hence $$ f(z):=z^2\sin\frac1z=\sum_{n=0}^{+\infty}\frac{(-1)^n}{(2n+1)!}\frac1{z^{2n-1}}=z-\frac{1/6}z+\dots\;\; $$

which allows us to conclude that $\operatorname{Res}(f,0)=-\frac16$.

$\endgroup$
2
  • $\begingroup$ Ah, that seems tricky for me. What was your motivation for that solution? $\endgroup$ – Rentop May 2 '15 at 9:18
  • $\begingroup$ I don't understand what do you mean by "motivation". $\endgroup$ – Joe May 2 '15 at 10:45

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.