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Cauchy-Goursat Theorem question

  1. Dec 23, 2013 #1
    I was reading about Cauchy-Goursat theorem and one step in the proof stumped me. It's the easier one, that is, Cauchy's proof that requires the complex valued function f be analytic in R, and f' to be continuous throughout the region R interior to and on some simple closed contour C. So that the contour integral around C is equal to zero.

    Also let f(z) = u(x,y) + i v(x,y).

    The proof used the hypothesis of Green's theorem which required the two real functions u and v and their first order partial derivatives on R to be continuous. I was thinking that if f is already analytic in R, wouldn't that already require u and v to be continuous in R and its first order partial derivatives? Since the condition for the differentiability of f requires it to be so? Doesn't that mean that the assumption that f' be continuous unnecessary?

    Edit: Oops, ok, I think I got it. I have confused myself with the theorem for 'sufficient condition for differentiability', which says that u and have to be continuous (and they have to obey the Cauchy-Riemann equations) in R. If such conditions are met by f, it implies its differentiability, but is the converse true?

    I might have skipped a lot of stuff about the theorem and the proof so my question might be confusing, but hopefully someone well acquainted with the topic would step in and help.
    Last edited: Dec 23, 2013
  2. jcsd
  3. Dec 24, 2013 #2
    Is my question too cluttered and messy? I'll revise my question if it is. Because I'd really appreciate it if someone could give their insight on this.
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