I can't find how to calculate path integrals of complex functions in the complex plane.
For example: $$\oint_{\mid z \mid =2}\frac{1-e^z+z}{z^3 (z-1)^2}dz$$
Mathematica Stack Exchange is a question and answer site for users of Wolfram Mathematica. It only takes a minute to sign up.
Sign up to join this communityFor this function:
f[z_] := (1 - E^z + z)/(z^3 (z - 1)^2)
there are no branch cuts in the complex plane therefore we simply use Cauchy integral theorem and the related formula of the complex residue, i.e. we sum up residues of the function $f$ in the circle $\mid z \mid =2$. Let's denote $$int = \oint_{\mid z \mid =2}\frac{1-e^z+z}{z^3 (z-1)^2}dz$$ Now we have:
int = 2 Pi I Total[ Residue[f[z], {z, #}] & /@ {0, 1}] // Simplify
I (-11 + 4 E) Pi
Alternatively we can parametrize z
over the given circle z -> 2 E^(I t)
:
(1 - E^z + z)/(z^3 (z - 1)^2) /. z -> 2 E^(I t)
(E^(-3 I t) (1 - E^(2 E^(I t)) + 2 E^(I t)))/(8 (-1 + 2 E^(I t))^2)
and d z -> 2 I E^(I t) d t
, now we have:
Integrate[(E^(-3 I t)(1 - E^(2 E^(I t)) + 2 E^(I t)))/( 8(-1 + 2 E^(I t))^2) 2 I E^(I t),
{t, 0, 2 Pi}]
I (-11 + 4 E) Pi
% // TraditionalForm
If a numerical answer is good enough you can just enter the path. As @Artes said it doesn't have to be the circle exactly.
NIntegrate[f[z], {z, 2 - 2 I, 2 + 2 I, -2 + 2 I, -2 - 2 I, 2 - 2 I}]
(* 0. - 0.398582 I *)
Check :
I (-11 + 4 E) Pi // N
(* 0. - 0.398582 I *)
Another suggestion from @Artes (thanks !) : one can use symbolic integration as well and
Integrate[f[z], {z, 2 - 2 I, 2 + 2 I, -2 + 2 I, -2 - 2 I, 2 - 2 I}] // FullSimplify
will reproduce his result.
Integrate
? Perhaps there is it somewhere, but couldn't find it.
$\endgroup$
z = Exp[I*t]
and havet
vary from 0 to 2*Pi $\endgroup$