# Laurent series(a really hard one!)

1. Oct 14, 2009

### oddiseas

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

((e^{iz})/((1+z²)))
Find a Laurent expansion valid for 0<|z-i|<1and use it to show that the derivative will leave you with the residue a₋₁.
So we have an annulus with its center at one of the singularities. Since the other singularity at z=-i is outside of the annulus it is not a problem, because the function is analytic. .
The problem is i have never found a laurent expansion for a function this complicated! with double poles!

2. Relevant equations

3. The attempt at a solution

((e^{iz})/((1+z²)²))=((e^{iz})/((1+z²)(1+z²)))=((e^{iz})/((i+z)(z-i)(i+z)(z-i)))=((e^{iz})/((i+z)²(z-i)²))
let w=z-i
z+i=w+2i
z=w+i
((e^{iz})/((i+z)²(z-i)²))=((e^{i(w+i)})/((w+2i)²(w)²))=((e^{iw})/(e(w+2i)²(w)²))=((e^{iw})/e)((1/(w²)))((1/(4[i-(-w)]²))

=((e^{iz})/4)*∑{n=0 to ∞}(-w)²ⁿ⁻²

I cant figure this question out! so if anyone can help

2. Oct 14, 2009

### HallsofIvy

Staff Emeritus
Write the function as
[tex]f(z)= \frac{e^{iz}}{z+i}\frac{1}{z- i}[/itex]
As you say, $e^{iz}/(z+ 1)$ is analytic at z= i and so can be written as a Taylor series there. Multiplying that by 1/(z-i) only decreases each power of (z- i) in the series by 1.