Cauchy's integral theorem in complex analysis
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The Cauchy's integral theorem states: Let U be an open subset of C which is simply connected, let f : U → C be a holomorphic function, and let γ be a rectifiable path in U whose start point is equal to its end point, then
∮γ [f(z)dz]=0.
Now I have a problem with this function e^z which is holomorphic in whole C. Taking γ as a circumference of unitary radius(γ=e^(jϑ)) by the theorem the integral should be null. On the other side
ro=1
g = @(theta) cos(theta) + 1i*sin(theta);
gprime = @(theta) -sin(theta) + 1i*cos(theta);
fun= @(z) exp(z);
q1 = integral(@(t) fun(g(t)).*gprime(t),0,2*pi)
resulting q1 = -1.2490e-16 - 9.9920e-16i Is it a calculus approximation or an error in the code?
P.s. Sorry for bad English it isn't my native language.
1 Comment
Torsten
on 25 Jan 2016
Don't you think the result is close enough to 0 :-) ?
Best wishes
Torsten.
Answers (1)
Diego Taglialatela
on 25 Jan 2016
0 votes
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