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!