I want to integrate the function
f[z]= z + Conjugate[z]
over a circle of radius 2 centered at the origin.
For the sake of stating something that I have tried:
Integrate[z + Conjugate[z], {z, 1, I, -1, -I, 1}]
seems to give me the integral around the square with vertices 1, I, -1, -I. Also I can get the integral on a line segment with:
Integrate[z + Conjugate[z], {z, 0, 1 + I}].
How do I integrate on a curve?
First we shall define 'the integration on a curve'. Traditionally, this is defined as integration of f.dl where dl is the length of a small part of the curve.
So, using t as a medium, we can explicitly write out the curve's function on a complex plane, here let's assume it's z=2 Exp[I t].
Then we can use t, a real number, as the integration variable, which make this problem significantly easier.
The final code is shown below:
z=2 Exp[I t];
f[z_]:=z+Conjugate[z];
Integrate[f[z] D[z,t],{t,0,2 Pi}]
result is:
8 I Pi
Hope this answer is helpful to you:)
Abs there? Which definition are you following?
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Abs, I thought the question was about integral of a real value over a curve, which requires a Abs.
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EDIT
It can get more complicated. Look at a contour given by
r[t_] := 1 + 2 Cos[t];
Original post
The result of the integral for a (reasonable) closed contour is just 2 I times the area enclosed by the contour.
Proof: the integral is
$$\int \left(z^*+z\right) \, dz$$
Letting
$$z=x+i y$$,
$$dz=dx+i dy$$
the integral becomes
$$\int 2 x (dx + i dy)=2 i \int dy* x - 2 \int dx *x $$
It can be easily shown that the second summand vanishes over a closed curve which completes the proof. QED.
As of version 13.3, ContourIntegrate is introduced:
Clear[z];
f[z_] := z + Conjugate[z];
ContourIntegrate[f[z], z ∈ Circle[{0, 0}, 2]]
(* 8 I π *)
Integrate function for the same purpose.
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ContourIntegrate: ContourIntegrate[E^(I z)/(z^3 + 1/8), z \[Element] RegularPolygon[6]].
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