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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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what is the meaning of $(u(x,y), -v(x,y))$ ? – Abdelmajid Khadari Mar 26 '12 at 21:30
1  
@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. – anon Mar 26 '12 at 21:36
up vote 3 down vote accepted

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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Why is the $v$ negative in $\vec{f}$? I thought that when a complex function say $x(t) + iy(t)$ was represented by a vector field it was $(x(t),y(t))$ rather than $(x(t),-y(t))$? – user26069 Mar 26 '12 at 22:21
    
@user26069: Apparently we are choosing to represent $f$ as a vector function a different way so that a meaningful line integral formula can be obtained. – anon Mar 26 '12 at 22:24
    
But why can we suddenly choose how to represent $f$ does that representation 'make sense' mathematically? (Sorry about all these questions!) – user26069 Mar 26 '12 at 22:34
    
@user26069: No problem with asking questions! Technically, $f$ and $\vec{f}$ are two different functions - they have different domains and codomains, but there is a sense in which they are, underneath the details, "the same" (almost, anyway), which is the motivation for calling them the same letter. The map $\mathbb{C}\to\mathbb{R}^2:x+iy\mapsto(x,y)$ is well-defined and invertible and smooth; $\tau:x+iy\mapsto(x,-y)$ is too. If we "forget" the complex structure of $\mathbb{C}$ (we must do so to move into a vector-calc setting) and view it as a real space, then our map is an isomorphism... – anon Mar 26 '12 at 22:52
    
Isomorphisms intuitively mean two structures are "the same." Thus, in a categorical sense, if $f:A\to A$ and $\tau:A\xrightarrow{\sim} B$ is an isomorphism, then $\tau\circ f\circ \tau^{-1}:B\to B$ is "the same" as $f$, only manifesting in a different context. Sorry if that was abstruse! – anon Mar 26 '12 at 22:53

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