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