Question

Determine the residue of e^zt/(z-2)^3

Answer 1

The function$f(z)=e^zt/(z-2)^3$ has only two special points: 2, $\infin$ (f has a singularity at z = 2, z=$\infin$ only). Therefore $Res(2,f)=-Res(\infin,f)$.

$Res(2,f)=c_{-1} -\text{ Laurent's coefficient}.$

$e^z=e^2+\frac{e^2}{1!}(z-2)+\frac{e^2}{2!}(z-2)^2+...$

Therefore Laurent's series for f is

$f(z)=t(… +\frac{e^2}{2!}(z-2)^{-1}+...)$.

Then

$Res(2,f)=t\frac{e^2}{2}, Res(\infin,f)=-t\frac{e^2}{2}.$

Finally, if z is a non singularity point for f, then $Res(z,f)=0.$