Solving Complex Line Integrals

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Josée Mallah
Josée Mallah 2021-3-21
评论： Bjorn Gustavsson 2021-3-22
Hello everyone!
How to define a circle (e.g. | z-1 |=3) as an integral path using Waypoints?
For example, C here is a square contour:
C = [1+i -1+i -1-i 1-i];
q2 = integral(fun,1,1,'Waypoints',C)
How to define C as a circle?
Or else, how could I define the same circle ( | z-1 |=3) instead of the unit circle in the code below?
g = @(theta) cos(theta) + 1i*sin(theta);
gprime = @(theta) -sin(theta) + 1i*cos(theta);
q1 = integral(@(t) fun(g(t)).*gprime(t),0,2*pi)
Thank you
  1 个评论
darova
darova 2021-3-21
THe question is unclear. Do you have any picture or something? What are you trying to integrate?

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Bjorn Gustavsson
Bjorn Gustavsson 2021-3-22
Wouldn't it be simplest to just multiply the radius of the polar representation of your unit-circle with the desired radius?
g = @(theta) 3*cos(theta) + 1i*3*sin(theta);
gprime = @(theta) -3*sin(theta) + 1i*3*cos(theta);
q1 = integral(@(t) fun(g(t)).*gprime(t),0,2*pi);
If you have some more complex curve sometimes those too can be more straightforwardly represented in polar coordinates. Then
You might also be lucky enough to have an analytical function - which makes integration around a closed loop in the complex plane that much easier.
HTH
  2 个评论
Josée Mallah
Josée Mallah 2021-3-22
I guess this is satisfactory enough! Ok I got how it works, thank you so much.
Bjorn Gustavsson
Bjorn Gustavsson 2021-3-22
Great.
It's just to remember that any single-variable parameterization of your perimeter is good enough as long as you have some way to calculate dl too.

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