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Complex Analysis Line Integral Question?
I'm having trouble evaluating the line integral around the curve y(t) = e^it, 0 <= t <= 2pi (unit circle) of Re(f) where f(z) = z. I substituted y(t) in for f which gave me
Re(y(t)) = Re(e^it) = Re(cos(t) + isin(t)) = cos(t)
and y'(t) = ie^it
so our integral is equivalent to the integral from 0 to 2pi of their product, namely cos(t)ie^it. Is this the correct procedure? Because at this point I tried to evaluate this integral by parts and got 0 but the answer should be i(pi).
Zo Maar, thanks a bunch! That certainly solved it =)
1 réponse
- Zo MaarLv 5il y a 1 décennieRéponse favorite
Your procedure is correct. It is not clear how you got zero in your answer.
exp(it) = cos(t) + i sin(t), and
cos(t)* i exp(it) = i cos^2 (t) - cos(t) sin(t).
When you integrate this expression over t from 0 to 2π, the term i cos^2 (t) = i [1/2 + cos(2t)/2] gives i π, while the second term cos(t) sin(t) = (1/2) sin(2t) drops out.
