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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!

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    @Matrix: Apparently OPs text/notes define $f$ as both $u+iv$ (complex function) and $(u,-v)$ (multivariable function), which was overloading the letter $f$ and confusing the OP.2012-03-26

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It would help if we wrote $f=u+iv$ and $\vec{f}=(u,-v)$ to distinguish between the two. Then

$\begin{array}{} (u+iv)(\dot{x}+i\dot{y}) & =(u\dot{x}-v\dot{y})+i(v\dot{x}+u\dot{y}) \\ & =(u,-v)\cdot(\dot{x},\dot{y})+i(u,-v)\cdot(\dot{y},-\dot{x}) \\ & =(\vec{f}\cdot\vec\tau)+i (\vec{f}\cdot \vec{n}). \end{array}$

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    @user26069: Yes, that's what I'm saying in my first comment to this answer.2012-03-26