Below is the proof that we did in lectures for the link between complex line integral and the line integral of a vector field.
If $f$ is the vector field associated to $f:\Omega \to \mathbb{C}$ by $f(x,y) = (u(x,y), -v(x,y))$ and $z(t) = x(t) + iy(t)$. Then $$\int_C f dz = \int_C f.\tau \, ds + i\int_Cf.n\,ds$$ where $\tau$ is the unit tangent and $n$ is the unit normal. The proof of this goes as such. $$\int_C f\,dz = \int_a^bf(z(t))\frac{dz}{dt}dt$$ $$= \int_a^b\left(u(x(t),y(t))\frac{dx}{dt} + iv(x(t),y(t))\frac{dy}{dt}\right)dt$$ $$+ i\int_a^b\left(u(x(t),y(t))\frac{dy}{dt} + iv(x(t),y(t))\frac{dx}{dt}\right)dt$$ $$ = \int_a^bf(x(t),y(t)).\left(\frac{dx}{dt},\frac{dy}{dt}\right) + i\int_a^bf(x(t),y(t)).\left(\frac{dy}{dt},-\frac{dx}{dt}\right)$$ $$= \int_C f.\tau \, ds + i\int_Cf.n\,ds \qquad \square.$$
But I don't understand how one gets from the first line to the second line of the proof? I'm trying to grapple with the concept of a complex function seemingly being represented by both $f(x,y) = (u(x,y), -v(x,y))$ and $f(x,y) = u(x,y) + iv(x,y)$ at the same time! Could anyone help with this? Thank you!
It would help if we wrote $f=u+iv$ and $\vec{f}=(u,-v)$ to distinguish between the two. Then
$$\begin{array}{} (u+iv)(\dot{x}+i\dot{y}) & =(u\dot{x}-v\dot{y})+i(v\dot{x}+u\dot{y}) \\ & =(u,-v)\cdot(\dot{x},\dot{y})+i(u,-v)\cdot(\dot{y},-\dot{x}) \\ & =(\vec{f}\cdot\vec\tau)+i (\vec{f}\cdot \vec{n}). \end{array}$$