I'm trying to put this in terms of e^(iz). Would this just be 2e^(i(theta)) where the angle ranges from 0 to pi? I'm trying to evaluate the integral over gamma of (z^2-1)dz
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Use z = 1 + 2e^(it) for t in [0, π].
I hope this helps!
Use z = 1 + 2e^(it) for t in [0, π].
I hope this helps!
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[1 + 2e^(it)]^2 - 1 = 4e^(2it) + 4e^(it)
Integrating this from 0 to π:
(1/i) [2e^(2it) + 4e^(it)] {for t = 0 to π}
= (1/i) [2 - 4] - (1/i) [2 + 4]
= 8i.
Integrating this from 0 to π:
(1/i) [2e^(2it) + 4e^(it)] {for t = 0 to π}
= (1/i) [2 - 4] - (1/i) [2 + 4]
= 8i.
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Did you use the formula integral f(z)dz = integral from a to b of f(gamma(t))*gamma'(t) dt to solve the above?
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