Complex contour integral proof

In summary, the "Complex contour integral proof" demonstrates the evaluation of integrals in the complex plane using techniques such as Cauchy's integral theorem and residue calculus. It involves the application of contour integration around singularities, allowing for the determination of integral values based on the residues at those points. This proof highlights the relationship between complex analysis and real integrals, showcasing the power of contour integration in simplifying complex problems.
  • #1
GGGGc
Homework Statement
How can I show that from the contour C_N (I’ve attached) that absolute value of cot(pi*z) is less than or equal to 0 everywhere on vertical parts of C_N and less than or equal to a value everywhere on the horizontal parts?
Relevant Equations
|a+b|<=|a|+|b|
|a-b|>=|a|-|b|
I’ve attached my attempt. I’ve tried to use triangle inequality formula to attempt, but it seems I got the value which is larger than 1. Which step am I wrong? Also, it seems I cannot neglect the minus sign in front of e^(N+1/2)*2pi. How can I deal with that?
 

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  • #2
You got as far as [tex]
|\cot \pi z | = \left|\frac{e^{2i\pi z} + 1}{e^{2i\pi z} - 1} \right| = \left|
\frac{e^{2i\pi x}e^{-2\pi y} + 1}{e^{2i\pi x}e^{-2\pi y} - 1}\right| [/tex] Why not go further? Multiply numerator and denominator by the complex conjugate of the denominator. Then you can easily write down the exact value of [itex]|\cot \pi z|^2[/itex], and at that point you can start trying to bound it on each side of the contour.

Or write [tex]
|\cot \pi z| = \frac{|e^{2i\pi x} + e^{2\pi y}|}{|e^{2i\pi x} - e^{2\pi y}|}[/tex] and then you can obtain an upper bound by maximizing the distance between [itex]e^{2i\pi x}[/itex] and [itex]-e^{2\pi y}[/itex] in the numerator and mimizing the distance between [itex]e^{2i\pi x}[/itex] and [itex]e^{2\pi y}[/itex] in the denominator.
 
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