residue(z*exp(1/z),z,0);

• Subject: residue(z*exp(1/z),z,0);
• From: Part Marty
• Date: Sat, 9 Jul 2011 23:51:51 +0900

Please tell me that.
 
(%i1)f:z*exp(1/z);
        residue(f,z,0);
        f2:taylor(f,[z,0,5,'asymp]);
(%o1) z*%e^(1/z)
(%o2) 0
(%o3)/T/ z+1+1/(2*z)+1/(6*z^2)+1/(24*z^3)+1/(120*z^4)+1/(720*z^5)+...
 
I think the residue of the expression "f" is 1/2. 
Why not ?
 
Function: residue (expr, z, z_0) 
Computes the residue in the complex plane of the expression expr when the variable z assumes the value z_0. The residue is the coefficient of (z - z_0)^(-1) in the Laurent series for expr.  		 	   		  

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