MHB Complex Valued Functions BV: John B. Conway Prop 1.3 Explained

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The discussion centers on understanding a specific aspect of Proposition 1.3 from John B. Conway's "Functions of a Complex Variable I." Participants seek clarification on the equality between two expressions involving the derivative of a complex-valued function, specifically how the sum of absolute values relates to an integral. It is clarified that while the derivative is not constant, the value at specific points, denoted as γ'(τ_k), is treated as a constant during integration. The conversation emphasizes the distinction between complex functions and complex numbers, confirming that the integral calculation is valid. This exchange highlights the nuances of complex integration and the interpretation of derivatives in this context.
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I am reading John B. Conway's book, "Functions of a Complex Variable I" (Second Edition) ...

I am currently focussed on Chapter IV: Complex Integration ... Section 1: Riemann-Stieljes Integral ... ...

I need help in fully understanding another aspect of the proof of Proposition 1.3 ...Proposition 1.3 and its proof read as follows:View attachment 7439
View attachment 7440
View attachment 7441
In the above text from Conway we read the following:

" ... ... Hence

$$
\int_a^b \lvert \gamma' (t) \rvert \ dt \le \epsilon + \sum_{ k = 1 }^m \lvert \gamma' ( \tau_k ) \rvert ( t_k - t_{ k - 1 } )$$$$ = \epsilon + \sum_{ k = 1 }^m \left\lvert \int_{ t_{ k-1 }}^{ t_k } \gamma' ( \tau_k ) dt \right\rvert $$ ... ..."
Can someone please explain exactly how/why $$\sum_{ k = 1 }^m \lvert \gamma' ( \tau_k ) \rvert ( t_k - t_{ k - 1 } ) = \sum_{ k = 1 }^m \left\lvert \int_{ t_{ k-1 }}^{ t_k } \gamma' ( \tau_k ) dt \right\rvert$$as is implied by the above quote from Conway ... ...?*** NOTE *** Seems as if Conway is treating $$\gamma' ( \tau_k )$$ as a constant ... but why ...?Help will be much appreciated ... ...

Peter
 
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It's because, for all $k$, $$\lvert \gamma'(\tau_k)\rvert(t_k - t_{k-1}) = \lvert \gamma'(\tau_k)(t_k-t_{k-1})\rvert = \left\lvert \int_{t_{k-1}}^{t_k} \gamma'(\tau_k)\, dt\right\rvert$$
 
Euge said:
It's because, for all $k$, $$\lvert \gamma'(\tau_k)\rvert(t_k - t_{k-1}) = \lvert \gamma'(\tau_k)(t_k-t_{k-1})\rvert = \left\lvert \int_{t_{k-1}}^{t_k} \gamma'(\tau_k)\, dt\right\rvert$$
Sorry to be slow Euge ...

... but I do not see why $$\lvert \gamma'(\tau_k)(t_k-t_{k-1})\rvert = \left\lvert \int_{t_{k-1}}^{t_k} \gamma'(\tau_k)\, dt\right\rvert$$Can you help further ...

hmm ... but maybe I guess that although $$\gamma' $$ is not a constant ... $$\gamma'( \tau_k)$$ is a constant, say $$\gamma'( \tau_k) = K$$ and so

$$\int_{t_{k-1}}^{t_k} \gamma'(\tau_k) dt = \int_{t_{k-1}}^{t_k}K dt = K ( {t_{k-1}} - {t_k} )
$$Is that correct?

Peter
 
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Since $\gamma'$ is a complex-valued function, $\gamma'(\tau_k)$ is a complex number, not a complex function as it appeared you were thinking. Your integral calculation is correct.
 
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