Homework StatementThis is actually on Wikipedia, but it doesn't show how you actually calculate the residues.
I want to calculate the residues for ((log(z))^2)/(1+z^2)^2. Wikipedia claims the sum of the residues is (-(pi/4) + (i*(pi)^2)/16 - (pi/4) - (i*(pi)^2)/16) = -pi/2
The Attempt at a SolutionThe singularities are at i and -i. Multiplying first by (z-i)^2 and taking the limit z ---> i, I get (log(i)^2)/4 = (log(e^(i*(pi/2))^2)/4 = -(pi^2)/16.
Then, multiplying by (z+i)^2 and taking the limit z ---> -i I get (log(e^-i*(pi/2))^2)/4 = -(pi^2)/16 again. Any help would be very much appreciated.