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Oh wait, I just noticed... and since for n ≠ 1, z ∈ Z, then every term reduces to zero, so the contour integral is zero... right?
Thanks for that, I did not know of this method. Is that the only way to show a function like this is holomorphic on C? Would I have to do this before I evaluate any contour integral?
Let so where and .
Since the partial derivatives are continuous for all and satisfy the Cauchy–Riemann equations and , is holomorphic on the complex plane.
Why is it holomorphic on C?
Then shouldn't it be 0 by Cauchy's integral theorem since is holomorphic on ?
C is some simple closed curve about 0.
What is C?
I was trying to find Let z = ln(t), then dz = (1/t)dt. This transforms the integral to Is this correct? And what would the graph of this look like? 