Expressing a complex contour integral in terms of vector integrals

Question

Let $z=x+iy$ and $f(z)=u(x,y)+iv(x,y)$ and let $C$ be a directed smooth curve. Define $\vec{f}=(u,v)$ , $d\vec{r}=(dx,dy)$ , $\vec{F}=(u,v,0)$ and $d\vec{R}=(dx,dy,0)$. What I want to do is, express $\int_C f(z)\,\textrm{d}z$ in terms of (not necessarily in terms of but somehow manipulating them) $\int_C \vec{f}\cdot d\vec{r}$ and $\int_C \vec{F}\times d\vec{R}$. What I did so far is to introduce $i$ to the vector fields and express the real part in terms of those terms. So, here is what I did \begin{align*} \int_C f(z)\,dz &= \int_C (u+iv)(dx+idy) \\ &=\int_C (u\,dx-v\,dy)+i(u\,dy+v\,dx) \end{align*} $$ \vec{f_z}=(u,iv)\,,\quad d\vec{r}_z = (dx,idy) \implies \int_C \vec{f_z}\cdot d\vec{r}_z = \int_C (u\,dx-v\,dy)\\ \implies \operatorname{Re}\left(\int_Cf(z)\,dz\right)=\int_C \vec{f_z}\cdot d\vec{r}_z\\[0.7cm]$$ \begin{align*} \int_C \vec{F}\times d\vec{R}&=\int_C(u,v,0)\times(dx,dy,0)\\ &=\int_C (0,0,u\,dy-v\,dx) = \left(0,0,\int_C(u\,dy-v\,dx)\right) \end{align*} So, I am stuck here, I tried a lot variations on the last integral but I couldn't succeed yet.

Answer 1

I figured it out with some help. \begin{align*} \int_C \overline{f(z)}\,dz &= \int_C (u-iv)(dx+i\,dy)\\ &=\int_C (u\,dx+v\,dy)+i(u\,dy-v\,dx)\\ &=\int_C (u\,dx+v\,dy) + i\int_C (u\,dy-v\,dx)\\ &=\left( \int_C \vec f\cdot d\vec r \right) + i \left(\hat k\cdot\int_C \vec F\times d\vec R\right) \end{align*}