0
$\begingroup$

I'm having trouble understanding exactly how to compute a complex line integral in $\mathbb{C}$. With my understanding of multivariable calculus, I view the line integral of a vector field $F: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ over some curve $C$ parameterized by $\gamma$ on $[a,b]$ as $$\int_{C} F \cdot dx = \int_{a}^{b}F(\gamma(t)) \cdot\gamma'(t)dt = \int_{a}^b F_1(\gamma(t))\gamma_1'(t) + F_2(\gamma(t))\gamma_2'(t)dt$$ where subscripts indicate component functions. According to my Professor for the Complex Variables course I am taking, this is not the same type of expansion we do on a line integral in $\mathbb{C}$. Instead, he claims that, if $F: \mathbb{C} \rightarrow \mathbb{C}$ and $\gamma$ is still some parameterization of a curve $G \subset \mathbb{C}$, then $$\int_{G} F \cdot dx = \int_{a}^{b}F(\gamma(t))\gamma'(t)dt$$ where I am to interpret $F(\gamma(t))\gamma'(t)$ as simple multiplication of complex numbers. Further confusing me, the textbook I am using (Fisher's Complex Variables) writes the result of Green's theorem as enter image description here

In which the left side appears to be consistent with my interpretation of line integrals from Multivariable calculus. What is going on here? How do I compute these line integrals?

EDIT: Here is an example from Fisher's that demonstrates what my Professor and Fisher think about line integrals enter image description here

$\endgroup$
  • $\begingroup$ Cutting ahead, you don't. If you have a holomorphic function the line contour integral around an closed contour equals 0. Otherwise the integral need only be evaluated at the points inside the where it fails to be analytic, regardless of the shape of the contour. You will learn that in a couple of weeks. In the mean time, you can use the techniques you learned in multi-variate calculus. $\endgroup$ – Doug M Jun 10 '16 at 1:03
  • $\begingroup$ My prof claims using these techniques are incorrect and the answers in the back of Fisher's seem to agree with him, what do I do? $\endgroup$ – Atreus Jun 10 '16 at 1:11
  • $\begingroup$ Do you have an example? $\endgroup$ – Doug M Jun 10 '16 at 1:54
  • $\begingroup$ Here i.imgur.com/69Ix3cc.png $\endgroup$ – Atreus Jun 10 '16 at 1:57
  • $\begingroup$ Sorry, I can't see that site from this computer. $\endgroup$ – Doug M Jun 10 '16 at 1:59
1
$\begingroup$

$\int_c f(z) dz = \int_\gamma f(\gamma(t))\gamma'(t) dt$

This is done just like a u-substitution.

And you can have some choices for $\gamma(t)$ i.e.

$\gamma(t) = 2 \cos t + 2i \sin t$ and $\gamma(t) = 2 e^{it}$ will both work in this case. In the example above they have used the second of those substitutions.

$f(z) = z^2 dz$ is a "holomorphic" function which is the complex equivalent to a conservative field in multivariate calculus. The integral depends only on the endpoints, and the choice of path does not matter.

Since integration is linear i.e. $\int f(z) + g(z) dx = \int f(z) dz +\int g(z) dx$ So, you can break this out if you like.

$\int_2^{2i} z^2 dz = \frac 13 z^3 = \frac 13 (2i)^3 - 8 = \frac 13 (-8i - 8)$

Finally $f(x + i y) = u(x,y) + i v(x,y)$

$\int_c f(z) dz = \int u dx - v dy + i\int u dy + v dx$

So, lets crank through this example using the last technique.

$u(x,y) = x^2 - y^2 - 3\sqrt{x^2 + y^2} + y\\ v(x,y) = 2xy\\ x = 2 \cos t\\ y = 2 \sin t\\ dx = -2 \sin t\\ dy = 2 \cos t$

$\int_c u(t) dx = \int_0^{\frac{\pi}{2}} (4\cos^2 t - 4\sin^2 t - 6 + 2 \sin t)(-2 \sin t) dt\\ \int (-16\cos^2 t \sin t + 20 \sin t -4 \sin^2 t)dt\\ \frac{16}{3}\cos^3 t - 20 \cos t -2t + 2\sin t\cos t|_0^{\frac{\pi}{2}}\\ -\frac{16}{3} + 20 - \pi $

$\int v(t) dy = \int_0^{\frac{\pi}{2}} 16 \sin t \cos^2 t dt\\ \frac {16}3$

The real part: $20 - \frac {32}3 - \pi = \frac {28}3 - \pi$

$\int_c u(t) dy = \int_0^{\frac{\pi}{2}} (4\cos^2 t - 4\sin^2 t - 6 + 2 \sin t)(2 \cos t) dt\\ \int_0^{\frac{\pi}{2}} (- 16\sin^2 t\cos t - 4\cos t + 4 \sin t\cos t) dt\\ \frac{-16}{3} \sin^3 t - 4 \sin t + 2 \sin^2 t |_0^{\frac{\pi}{2}}\\ \frac{-16}{3}-2$

$\int_c v(t) dx = \int_0^{\frac{\pi}{2}} -16\sin^2 t \cos t dt\\ \frac{-16}{3}$

the imaginary part = $\frac{-32}3 - 2 = \frac{-38}3$

Regarding Green's theorem, that only applies to closed contours.

$\endgroup$
  • $\begingroup$ So, just for certainty, the left hand statement of green's theorem (that I posted the screenshot of) is not equal to the line integral right? $\endgroup$ – Atreus Jun 11 '16 at 17:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.