Let $f : \mathbb C → \mathbb C$ be a function of class $C^1$ (not necessarily holomorphic); write $f = u + iv$.
Let $Ω ⊂ C$ be a domain with boundary $bΩ = C$, where $C$ is a simple, closed, piecewise differentiable curve.
Define the planar vector fields $F_1 = ui − vj$,
$F_2 = vi + uj$.
(a) Show that $\int_C{f(z)dz} = \int_C{F1 · dr} +i \int_C{F2 · dr} $.
(b) Use Green’s theorem to show that $\int_C{f(z)dz} = 2i \int \int_Ω {\frac{∂f} {∂\bar{z}}} dxdy$.
(c) What can we conclude if $f$ is $C^1$ and complex differentiable?
My attempts:
a) I know that $\int_C {F1 · dr}$ = $\int_C ∇f_1. dr$, so if I want to find the potential function $f_1$, I say that $f_{1x} = u(x, y) $ and so $f_1 = \int{u(x, y)}dx = h(x, y) + g(y) $ where $h(x, y)$ is a function in $x$ and $y$ and $g(y)$ is a function in $y$. So $f_{1y} = h_y(x, y) + g'(y) $
And I have $f_{1y} = - v(x, y) $, thus $-v(x, y) = h_y(x, y) +g'(y) $. Hence $g'(y) = -v(x, y) - h_y(x, y) $, so $g(y) = - \int{v(x, y)}dy - \int{h_y(x, y)}dy $
Then $f_1 = - \int{v(x, y)} dy$
I do the same procedure as before to find $f_2 = \int{u(x, y)} dy$
How do I continue from there to prove the equality?
b) I know Green's theorem is $\int_C {Ldx + Mdy} = \int \int_D (\frac{\partial M} {\partial x} - \frac{\partial L}{\partial y}) dxdy$.
How do I proceed from here?
c) I don't really know what to do.
Any help please in the 3 parts? Thank you