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futurebird

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**Complex analysis: having partials is the same as being "well defined?"**

My professor proved this theorem in class and I don't know if I even wrote it down correctly in my notes. I don't have access to the book so I need to know if this makes sense. Here is the theorem:

Under these conditions:

[tex]R[/tex] is a simply connected region in [tex]\mathbb{C}[/tex] and

p,q : [tex]R \rightarrow \mathbb{R}[/tex] continuous.

[tex]\sigma - rect[/tex] is a rectangle in [tex]\mathcal{R}[/tex].

The existence of a function U on [tex]R[/tex] such that:

[tex]\frac{\partial U}{\partial x}=p, \frac{\partial U}{\partial y}=q[/tex]

[tex]\Longleftrightarrow \displaystyle\oint_{\sigma - rect} pdx + qdy =0[/tex]

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QUESTIONS:

1. My professor said that having the partials for p and q is the same thing as saying that U is well defined. Why is this the case?

2. Then he said U would be well defined if it was path independent, well, if we know it's path independent then,

*of course*[tex]\displaystyle\oint_{\sigma - rect} pdx + qdy =0[/tex]. So, am I right to think of path independence as a consequence of this proof, not a condition? (I know this seems obvious, but I keep losing track of what we "know" when we start the proof and what we are trying to prove so, I want to get this absolutely clear!)

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