0
$\begingroup$

I was asked to find a Laurent-series expansion for $f(z) = z^{-1} \sinh(z^{-1})$ about the point 0, and to classify the singularity at 0.

Now, that wasn't too bad - with some manipulation of the expansion for $sinh(z)$, I got the following;

$$z^{-1} \sinh(z^{-1}) = \sum_{n=0}^{\infty} \frac{1}{(2n+1)!} (\frac{1}{z})^{2n+2} = \sum_{n=2}^{\infty} \frac{1}{(2n-1)!} (\frac{1}{z})^{2n}$$

Now, to classify the singularity at 0, I wrote out the first few terms of the expansion;

$$z^{-1} \sinh(z^{-1}) = \frac{1}{z^2} + \frac{1}{6z^4} + \frac{1}{120z^6} + ...$$

And noted that the function has negative powers of all orders, which means that 0 is an essential singularity of the function.

Is it correct to say this?? Is there a stronger proof of this??

As a bit of an additional exercise for myself, I wanted to determine the residue at this point. Using an example from my textbook related to determining the residue of $exp(\frac{1}{z^2})$ about the point 0, I kind of assumed that the residue would be equal to zero, as there are no $\frac{1}{z}$ terms within my series expansion for sinh, so the residue must be zero?? Again, is there a stronger way to prove this, if it's a correct statement??

$\endgroup$
  • $\begingroup$ seems okay to me $\endgroup$ – Santosh Linkha Jun 1 '14 at 5:15
  • $\begingroup$ Looks correct. :-) $\endgroup$ – Bobby Ocean Jun 2 '14 at 1:32

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.