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Math Help - Fundamental Theorem of Complex Calculus

  1. #1
    Senior Member slevvio's Avatar
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    Fundamental Theorem of Complex Calculus

    Hello everyone, I was having trouble with the proof of this and I was wondering if anyone could help me!

    Let f:S \rightarrow \mathbb{C} be analytic (where S is an open set), and such that if \gamma : [a,b] \rightarrow S is a piecewise continuously differentiable path in S, then if f' is continuous on \gamma^* (which is always true anyway), and  z_1:= \gamma (a), z_2:= \gamma (b),

    then \int_{\gamma} f'(z)dz = f(z_2)-f(z_1)

    PROOF: Write \gamma as \gamma_1 + \gamma_2 + \ldots + \gamma_n, where each \gamma_i : [t_{i-1}, t_i] \rightarrow S is a simple smooth path, with \gamma_i (t_i) = \gamma_{i+1} (t_i), \gamma_1 (t_1) := \gamma(a), \gamma_n (t_n) := \gamma (b).

    Then \int_{\gamma_i} f'(z) dz = \int_{t_{i-1}}^{t_i} f '(\gamma_i (t) ) \gamma_i ' (t) dt = \int_{t_{i-1}}^{t_i} \dfrac{d}{dt} f ( \gamma_i (t) ) dt = now here I have a problem because it says this line is equal to

     <br />
\left[ f(\gamma_i (t)) \right]_{t_{i-1}}^{t^i}<br />
by the fundamental theorem of calculus for REAL functions. however f \circ \gamma_i is not real valued is it? so why can i use it here?
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  2. #2
    MHF Contributor

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    Just apply the fundamental theorem of calculus to the real and imaginary parts of the function. Indeed, \int \phi(t)dt=\int {\rm Re}(\phi(t))dt+i\int {\rm Im}(\phi(t))dt for \phi:\mathbb{R}\to\mathbb{C} (and a function is differentiable iff both its real and imaginary parts are). Similarly, most theorems like integration by part or change of variables adapt seamlessly to complex-valued functions.
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  3. #3
    Senior Member slevvio's Avatar
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    ah I see this now! thanks very much
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