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A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 4.32, Cor 8.27

Question 1. Should the following 2 statements in the textbook have an assumption that the path $\gamma$ in question is positively oriented?

Question 2. Are there ways to forego assuming $\gamma$ is positively oriented? Eg 'If $\gamma$ is simple, piecewise smooth and closed but not positively oriented, then $-\gamma$ is' or something. This may be a Calculus III issue.

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2 statements:

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  1. (Cor 8.27) Suppose $f$ is a function that is holomorphic in $A = \{R_1 < |z-z_0| < R_2\}$ with Laurent series $$f(z) = \sum_{k=-\infty}^{\infty} c_k (z-z_0)^k$$ If $\gamma$ is any simple, closed, piecewise smooth, path in $A$ s.t. $z_0 \in int(\gamma)$, $$\int_{\gamma} f = 2\pi i c_{-1}$$
  1. (Exer 4.32) Show that the corollary (Cor 4.20) to Cauchy's Thm (Thm 4.18) is a corollary to Cauchy's Integral Formula (Formula 4.27) if $\gamma$ is simple.
  • Reason: Cauchy's Integral Formula (Formula 4.27) assumes $\gamma$ is positively oriented while Cor 4.20 doesn't.
BCLC
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    The statement of 8.27 makes no sense for several reasons. Did you state it exactly as it appeared in the book? If yes: You need a better book. If no: You should be much more careful - the little changes you made that meant the same thing really don't mean the same thing at all. – David C. Ullrich Aug 11 '18 at 03:15
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    In particular (i) saying $z_0\in A$ and then later saying $\forall z_0[\dots]$ makes no sense logically (ii) If $z_0\in A$ and $c_{-1}$ is the coefficient in the Laurent series for $f\in H(A)$ centered at $z_0$ then $c_{-1}=0$, because that Laurent series is a power series! I tend to suspect that $z_0$ is actually the center of $A$ - if so then you should say so; note that if $z_0$ is the center of $A$ then $z_0\notin A$. – David C. Ullrich Aug 11 '18 at 03:19
  • @DavidC.Ullrich thanks! Screenshots are apparently bad, and it was too time consuming to do word for word, at least in the short run because I guess it's more time consuming in the long run having to edit or explain. – BCLC Aug 11 '18 at 05:15
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    Right. I don't know if you noticed, but you didn't quite answer my question. Does the book say exactly what you wrote or not? If not you really should change the question to say what the book says. Would make the question much better. Also quoting someone incorrectly is a bad bad thing! The authors would be very unhappy with you if they knew you claimed the book contained that sort of nonsense. – David C. Ullrich Aug 11 '18 at 15:03
  • @DavidC.Ullrich not the same thanks for the reminder. I'll edit later – BCLC Aug 11 '18 at 16:48
  • @DavidC.Ullrich Ah ok I see the problem. I believe it's supposed to be 's.t. z_0 \in \int(\gamma)$' not '$\forall z_0 \in int(\gamma)$' – BCLC Aug 12 '18 at 06:49
  • A day later and you still can't be bothered to tell me whether it really says $z_0\in A$ or, as would make more sense, that $z_0$ is the center of $A$. Wow. – David C. Ullrich Aug 12 '18 at 14:50
  • @DavidC.Ullrich centre as I edited...? – BCLC Aug 13 '18 at 02:45
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    Fine. Much better - sorry about reading that comment and not looking for changes to the answer. (Didn't have time, heh...) – David C. Ullrich Aug 13 '18 at 02:56
  • @DavidC.Ullrich thanks! – BCLC Aug 13 '18 at 12:07

1 Answers1

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If $-\gamma$ is the reverse of contour $\gamma$, $\oint_{-\gamma} f = - \oint_\gamma f$. So the formula of Cor. 8.27 can't be true for both $\gamma$ and $-\gamma$: you do need to assume $\gamma$ is positively oriented.

On the other hand, Cor. 4.20 says $\oint_\gamma f = 0$: if that's true for $\gamma$, it's also true for $-\gamma$, so this does not need to assume positively oriented.

Robert Israel
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