- #1
KNUCK7ES
- 13
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Homework Statement
i have the closed ∫ (cos(z))/(3z-π)) dz
Homework Equations
i presume this works closed ∫ f(z)/(z-w) dz = f(w)2πi
The Attempt at a Solution
i think the answer is cos(pi)*2πi
KNUCK7ES said:i can't believe i didn't factor. now i get 1/3*pi*i
KNUCK7ES said:1. evaluate each integral around the counter-clockwise circle lzl =1 repeat for lzl=2.
(a) ∮ cos(z)/(3z-pi) dz, (b) ∮ z /((9-z^2)*(2z-i)) dz
thats all it says. i think that the pole may be not be significant
KNUCK7ES said:so that would mean that the integrand is analytic in the interior and on the contour, so it would equal 0?
KNUCK7ES said:so being aware of the poles, in the case of (b) we have poles at z=-3,3,i/2. but the only one that matters is the i/2 because it is the only one inside the contour line. my only problem is now trying to get ∮ z/((9-z^2)(2z-i)) dz into the form f(w) ∮ 1/(z-w) dz
KNUCK7ES said:i did both, and the partial fractions is a little more ugly. but where i became stuck, is that i began to wonder if i had to do it twice the other one being where f(z) = z/(2z-i)
Dick said:I'm not sure where the stuckness comes in. f(z)=z/(9-z^2) is analytic inside of both |z|=1 and |z|=2, isn't it? Just do f(z)/(2z-i). Just apply your theorem to that.
KNUCK7ES said:actually upon further thinking of my question, it would have to have bounds ∮ 1/(z^2-z+2) dz for lzl=1 would be 0 but if i were to extend the radius out to lzl =2 then it gets more interesting
A complex contour integration problem involves integrating a complex-valued function along a path in the complex plane, rather than just a real-valued function along a line in the real plane. This type of problem is often encountered in mathematics, physics, and engineering.
A complex contour integration problem is typically solved using the Cauchy integral theorem and Cauchy integral formula. These mathematical tools involve representing the complex-valued function as a series of real and imaginary parts, and then evaluating the integral using techniques such as residue calculus.
Complex contour integration is important in science because it allows for the evaluation of complex-valued functions and integrals, which often arise in physical and mathematical models. This type of integration also has applications in fields such as signal processing, fluid dynamics, and quantum mechanics.
One of the main challenges in solving complex contour integration problems is choosing the appropriate contour or path in the complex plane. This requires a deep understanding of the function being integrated and its singularities. Another challenge is accurately evaluating the integral, which can involve complex calculations and may require the use of advanced mathematical techniques.
Yes, complex contour integration has many real-world applications in fields such as engineering, physics, and economics. It can be used to solve problems involving complex-valued functions, such as calculating electric fields in electromagnetics or fluid flow in fluid dynamics. It is also used in data analysis and signal processing to extract information from complex-valued signals.