Question 1
I want to prove that given $ f:\mathbb{C}\to\mathbb{C} $ and rectangle $ R $ with boundary $\gamma $ anti-clockwise, the line integral over $ f $ would be $ 0 $.
I'm pretty sure with my way but I cannot find my mistake. Here's what I have tried:
Assume that $ f\left(x,y\right)=\left(u\left(x,y\right),v\left(x,y\right)\right) $ (we will recognize $\mathbb{C} $ with $\mathbb{R}^2$, and assume $ \gamma\left(t\right)=\left(x\left(t\right),y\left(t\right)\right),\thinspace\thinspace\thinspace a\leq t\leq b $
Then,by definition we have:
$$ \intop_{\gamma}fdz=\intop_{a}^{b}f\left(\gamma\left(t\right)\right)\cdot\dot{\gamma}\left(t\right)dt $$
$$ =\intop_{a}^{b}\left[\left(u\left(x\left(t\right),y\left(t\right)\right),v\left(x\left(t\right),y\left(t\right)\right)\right)\right]\cdot\left[\left(\dot{x}\left(t\right),\dot{y}\left(t\right)\right)\right]dt $$
$$ =\intop_{a}^{b}\left[u\left(x\left(t\right),y\left(t\right)\right)\dot{x}\left(t\right)+v\left(x\left(t\right),y\left(t\right)\right)\dot{y}\left(t\right)\right]dt $$
$$ =\intop_{\gamma}u\left(x,y\right)dx+v\left(x,y\right)dy\underset{\text{Green's Theorem}}{=}\intop_{R}\left(v_{x}-u_{y}\right)dxdy $$
I used Gree'ns theorem as it stated in wikipedia
What I actually need in those parenthesis is $ \left(u_{x}-v_{y}\right) $ so I can use Cauchy-Riemann equations and claim that the integrand is $0$, but seems like I was wrong somewhere
Any help and clarifications would be very helpful and I would highly appreciate it.
Thanks in advance.
We regard $f \colon \mathbb{R}^2 \to \mathbb{C}$. Your first one is just not the definition of contour integration. It gives you a real number, so is definitely not correct. Your second one is correct. The first integral you computed is (with $f = u + iv$) $$\int_{\partial R}u\,dx + v\,dy,$$ while the definition of the contour integral is $$\int_{\partial R}f\,dz := \int_{\partial R}f(dx + i\,dy) = \int_{\partial R}f\,dx + if\,dy.$$ Green's theorem applies to functions from $\mathbb{R}^2$ to $\mathbb{C}^2$ too (this follows easily from the real version of Green's theorem), so applying Green's theorem to $(f, if)$ gives $$\int_{\partial R}f\,dx + if\,dy = \int_{R}(if_x - f_y)\,d(x, y).$$ If $f$ is holomorphic, then $if_x - f_y = 0$, which yields your result.
Edit: The contour integral is defined as the line integral of the vector field $(f, if)$, a complex valued vector field. The integral of such a vector field is defined componentwise. That means it is really two real integrals. We can expand to see them \begin{align} f\,dz &= (u + iv)\,(dx + i\,dy) \\ &= u\,dx - v\,dy + i(u\,dy + v\,dx). \\ \end{align} So $$\int_{\partial R}f\,dz = \int_{\partial R}(u\,dx - v\,dy) + i\int_{\partial R}(v\,dx + u\,dy).$$