Answer to Question #223821 in Chemical Engineering for Lokika

Question

The radius of convergence of the Taylor series expansion of the functiorf(z)=3z+5/(z+1)^{2} (z-2) about z=1 is

Answer 1

To find radius of convergence of the function f(z) = $f(z)=\frac{3z+5}{(z+1)^{2} (z-2)}$ about z=1

We know that for a meromorphic function, the radius of convergence of the Taylor series centred at some point is just the distance from the closest singularity. 

$(z+1)^2 (z-2)$ vanishes at z=-1 and z =2 since the closest point from z =1 is z =2 and the distance is 1. 

Hence radius of convergence will be 1.