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A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Ch5.2

2 Questions about Cor 5.8 and Cor 5.9 (*)

Question 1. Can we prove Cor 5.9 using Cor 5.8?

Question 2. Can we prove Cor 5.9 without using Cor 5.8?

My proof for either starts out the same:

Consider two paths $\gamma_1, \gamma_2 \subset G$ that are piecewise smooth and have the same start and end points. Denote $-\gamma_2 \subset G$ as $\gamma_2$ passed in the reverse direction. Denote $\gamma_1 \wedge -\gamma_2 \subset G$ as the path that starts at the start of both $\gamma_1$ and $\gamma_2$ and passes $\gamma_1$ until the end of both $\gamma_1$ and $\gamma_2$, w/c is equivalent to the start of $-\gamma_2$ and then passes $-\gamma_2$ until the end of $-\gamma_2$, w/c is equivalent to the start of both $\gamma_1$ and $\gamma_2$. Observe that $\gamma_1 \wedge -\gamma_2 \subset G$ is a closed and piecewise smooth path and thus

$$0 \stackrel{(**)}{=} \int_{\gamma_1 \wedge -\gamma_2} f := \int_{\gamma_1} f + \int_{-\gamma_2} f := \int_{\gamma_1} f - \int_{\gamma_2} f \implies \int_{\gamma_1} f = \int_{\gamma_2} f$$

This shows that $\forall \gamma \subset G$ piecewise smooth, $\int_{\gamma} f$ has the same value because $\forall \gamma_1, \gamma_2 \subset G$ piecewise smooth with the same start and end points as $\gamma$, $\int_{\gamma_1} f = \int_{\gamma_2} f$.

$$\therefore, \int_{\gamma} f \ \text{is path independent.}$$

QED

(**) The justification for this I believe can be done using or without using Cor 5.8. I moved my attempts to an answer.


(*)

(Cor 5.8) If $f$ is holomorphic function on a simply-connected region $G \subseteq \mathbb C$, then $f$ has an antiderivative.

(Cor 5.9) If $f$ is a holomorphic function on a simply-connected region $G \subseteq \mathbb C$, then $\forall \gamma \subset G$ piecewise smooth, $\int_{\gamma} f$ is path independent.

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  • 2
    $\begingroup$ The opposite of "without using" is not "with using"; it's just "using". $\endgroup$ – Kenny Lau Aug 12 '18 at 7:36
  • $\begingroup$ @KennyLau Emphasis? I think it's like saying 'nowhere holomorphic and nowhere differentiable' Thanks $\endgroup$ – BCLC Aug 12 '18 at 7:39
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    $\begingroup$ Nobody says "with using", even for emphasis. $\endgroup$ – Kenny Lau Aug 12 '18 at 7:40
  • $\begingroup$ @KennyLau ayt edited thanks $\endgroup$ – BCLC Aug 12 '18 at 7:41
  • $\begingroup$ @CalvinKhor Thanks! ^-^ $\endgroup$ – BCLC Aug 13 '18 at 9:55
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Question 1.

Pf of Cor 5.9 using Cor 5.8:

Consider two paths $\gamma_1, \gamma_2 \subset G$ that are piecewise smooth and have the same start and end points. Denote $-\gamma_2 \subset G$ as $\gamma_2$ passed in the reverse direction. Denote $\gamma_1 \wedge -\gamma_2 \subset G$ as the path that starts at the start of both $\gamma_1$ and $\gamma_2$ and passes $\gamma_1$ until the end of both $\gamma_1$ and $\gamma_2$, w/c is equivalent to the start of $-\gamma_2$ and then passes $-\gamma_2$ until the end of $-\gamma_2$, w/c is equivalent to the start of both $\gamma_1$ and $\gamma_2$. Observe that $\gamma_1 \wedge -\gamma_2 \subset G$ is a closed and piecewise smooth path and thus by $\color{blue}{(1)}$ below,

$$0 \stackrel{\color{blue}{(1)}}{=} \int_{\gamma_1 \wedge -\gamma_2} f := \int_{\gamma_1} f + \int_{-\gamma_2} f := \int_{\gamma_1} f - \int_{\gamma_2} f \implies \int_{\gamma_1} f = \int_{\gamma_2} f$$

This shows that $\forall \gamma \subset G$ piecewise smooth, $\int_{\gamma} f$ has the same value because $\forall \gamma_1, \gamma_2 \subset G$ piecewise smooth with the same start and end points as $\gamma$, $\int_{\gamma_1} f = \int_{\gamma_2} f$.

$$\therefore, \int_{\gamma} f \ \text{is path independent.}$$

QED Cor 5.9 using Cor 5.8

Pf $\color{blue}{(1)}$:

We can use a corollary (Cor 4.13) to the complex analogue of the Fundamental Theorem of Calculus Part II (Thm 4.11) because

  • $f$ has an antiderivative on $G$ by Cor 5.8, w/c we can apply $\because f$ is holomorphic on a simply-connected region, by assumption

  • $f$ is continuous on an open subset $G \because f$ is holomorphic on $G$ by assumption,

  • $G$ is open $\because G$ is a simply-connected region by assumption and

  • $\gamma_1 \wedge -\gamma_2$ is piecewise smooth and closed $\because \gamma_1$ and $ -\gamma_2$ are too $\because \gamma_1$ and $\gamma_2$ are too.

  • Antiderivatives can be defined on open disconnected subsets and need not be only for regions as the book defines.

QED $\color{blue}{(1)}$

Question 2.

Pf of Cor 5.9 without using Cor 5.8:

Consider two paths $\gamma_1, \gamma_2 \subset G$ that are piecewise smooth and have the same start and end points. Denote $-\gamma_2 \subset G$ as $\gamma_2$ passed in the reverse direction. Denote $\gamma_1 \wedge -\gamma_2 \subset G$ as the path that starts at the start of both $\gamma_1$ and $\gamma_2$ and passes $\gamma_1$ until the end of both $\gamma_1$ and $\gamma_2$, w/c is equivalent to the start of $-\gamma_2$ and then passes $-\gamma_2$ until the end of $-\gamma_2$, w/c is equivalent to the start of both $\gamma_1$ and $\gamma_2$. Observe that $\gamma_1 \wedge -\gamma_2 \subset G$ is a closed and piecewise smooth path and thus by $\color{red}{(2)}$ below,

$$0 \stackrel{\color{red}{(2)}}{=} \int_{\gamma_1 \wedge -\gamma_2} f := \int_{\gamma_1} f + \int_{-\gamma_2} f := \int_{\gamma_1} f - \int_{\gamma_2} f \implies \int_{\gamma_1} f = \int_{\gamma_2} f$$

This shows that $\forall \gamma \subset G$ piecewise smooth, $\int_{\gamma} f$ has the same value because $\forall \gamma_1, \gamma_2 \subset G$ piecewise smooth with the same start and end points as $\gamma$, $\int_{\gamma_1} f = \int_{\gamma_2} f$.

$$\therefore, \int_{\gamma} f \ \text{is path independent.}$$

QED Cor 5.9 without using Cor 5.8

Pf $\color{red}{(2)}$:

Use a corollary (Cor 4.20) to Cauchy's Thm (Thm 4.18), w/c we can apply because:

  • $G$ is a region $\because G$ is a simply-connected region,

  • $\gamma_1 \wedge -\gamma_2 \sim_G 0 \because \gamma_1 \wedge -\gamma_2$ is closed, and closed paths are $G$-contractible if $G$ is a simply-connected region, by definition of a simply-connected region and

  • $\gamma_1 \wedge -\gamma_2$ is piecewise smooth and closed $\because \gamma_1$ and $ -\gamma_2$ are too $\because \gamma_1$ and $ \gamma_2$ are too.

QED $\color{red}{(2)}$

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