Please check find residue problem

[ognik]

Hi - I get a different answer from the book, but please also review for correct mathematical language & notation ...

find residues for $ f(z) = \frac{sin(\frac{1}{z}) }{z^2 + a^2} $ There are 2 simple poles, $ \pm ia $ ; also I note that $ {z}_{o}^2 = -a^2 $ which proves useful for simplifying.

$ Res[f, ia] = \lim_{{z}\to{{z}_{0}}}\frac{(z - {z}_{0})sin(\frac{1}{z})}{z^2 - {z}_{0}^2} =
\lim_{{z}\to{{z}_{0}}} \frac{sin(\frac{1}{z})}{z + {z}_{0}} = \frac{sin(\frac{1}{ia})}{2ia} $

Now $ sin(i\theta) = i sinh(\theta) $, so $ Res[f, ia] = \frac{sinh(\frac{1}{a})}{2a} = Res[f, -ia] $

But the book says it should be $ - \frac{sinh(\frac{1}{a})}{2a} $ ?

 

[Fernando Revilla]

But the book says it should be $ - \frac{sinh(\frac{1}{a})}{2a} $ ?

Follow the steps, and try to find your mistake when using $z_0^2=-a^2$
$$\text{Res}[f, ia] = \lim_{{z}\to{ ia}}\frac{(z - ia)\sin \left(\frac{1}{z}\right)}{(z-ia)(z+ia)} = \lim_{{z}\to{ia}} \frac{\sin\left(\frac{1}{z}\right)}{z + ia} = \frac{\sin\left(\frac{1}{ia}\right)}{2ia}$$ $$=\frac{\sin\left(-i\frac{1}{a}\right)}{2ia}=\frac{-\sin\left(i\frac{1}{a}\right)}{2ia}=\frac{-i\sinh\left(\frac{1}{a}\right)}{2ia}=-\frac{\sinh\left(\frac{1}{a}\right)}{2a}.$$

 

[ognik]

Yes, just a silly mistake, thanks Fernando