Thread: Finding the residue of the following function

1. Finding the residue of the following function

Hi all, I have the following function $f(z) = \frac{1}{(z^2 + z + 1)^2}$ and I need to find the residue at the double pole $z = (-1)^{\frac{2}{3}}$. Any tips? I'm hoping to not have to use the Laurent expansion for this. I know that $(z^2 + z + 1) = \frac{(1-z)}{(1-z^3)}$. I'm aware of a limit formula for higher order poles but I don't know how to manipulate the function to get it in that form. Maybe that's not even needed. Thanks!

2. Re: Finding the residue of the following function

Originally Posted by director
Hi all, I have the following function $f(z) = \frac{1}{(z^2 + z + 1)^2}$ and I need to find the residue at the double pole $z = (-1)^{\frac{2}{3}}$. Any tips? I'm hoping to not have to use the Laurent expansion for this. I know that $(z^2 + z + 1) = \frac{(1-z)}{(1-z^3)}$. I'm aware of a limit formula for higher order poles but I don't know how to manipulate the function to get it in that form. Maybe that's not even needed. Thanks!
Note that the (double) roots of f(z) are $\frac{-1\pm i\sqrt{3}}{2}$ (which are equivalent to $(-1)^{2/3})$. Set $z_1=\frac{-1+ i\sqrt{3}}{2}$ and $z_2=\frac{-1- i\sqrt{3}}{2}$, then $Res(f,z_1) = \frac{1}{(n-1)!} \lim_{z->z_1}\frac{d^{n-1}}{dz^{n-1}}(z-z_1)^2 f(z)$ where n=2. Similarily for $z_2$.