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Apr3-12, 09:21 PM
P: 19
Complex integral over a circle

Okay so third times a charm, i figured out that it is not f^(m-1), but f with derivatives equal to m-1.

So since my m=1 for this, i do not have to take a derivative but only evaluate it at 3/2 over !0.

so my answer is pi * i * 3/2 * e^(3/2)