Laurent series for f(z) = z^2 sin(pi (z+1)/z); 0<|z|<infinity. ?
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May 31st 2012, 08:00 AM
mariusrb
Laurent series for f(z) = z^2 sin(pi (z+1)/z); 0<|z|<infinity. ?
Laurent series for f(z) = z^2 sin(pi (z+1)/z); 0<|z|<infinity.