Answer to Question #223821 in Chemical Engineering for Lokika

Question #223821

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


1
Expert's answer
2021-08-12T07:28:13-0400

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.


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