# How do i get sin(z)/2i = 1/4(e - 1/e) ? Resides at Poles question?

1. Jun 6, 2007

### laura_a

1. The problem statement, all variables and given/known data

Find Poles & eval resides at those points for

f(z) = sin(z)/(z^2+1)

2. Relevant equations

I'm using Res f(z) = [g^m-1 (z_0)] / (m-1)!

3. The attempt at a solution

So I worked out the f(z) = 1/(z-i) so g(z) = sin(z) /(z+i) for my first one did the sub for z=i and got sin(i)/2i

But the working out (i'm working on past papers so I have the worked solutions, just not very detailed) The next line shows;

= 1/4(e-(1/e))

What one earth don't I know about sin that gets to this point??

Last edited: Jun 6, 2007
2. Jun 6, 2007

### mjsd

For simple pole at $$z_0$$, the residue of f(z) is given by
$$\text{Res}\left[f(z), z_0\right]= \lim_{z\rightarrow z_0} (z-z_0) f(z)$$

eg.
$$\text{Res}\left[\frac{1}{(z^2+1)}, i \right]= \lim_{z\rightarrow i} \frac{z-i}{z^2+1} =\lim_{z\rightarrow i} \frac{1}{z+i}=\frac{1}{2i}$$

now, looks like you have got it right so far with sin(i)/2i, all it is need to get that next line is just by using
$$\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$$