Complex contour integral proof

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The discussion focuses on a complex contour integral proof where the user struggles with applying the triangle inequality and managing a negative sign in their calculations. They express confusion over obtaining a value greater than 1 and seek guidance on correcting their approach. Suggestions include multiplying the numerator and denominator by the complex conjugate to simplify the expression for |\cot πz| and finding its exact value. Additionally, an alternative method is proposed to derive an upper bound by analyzing the distances in the numerator and denominator. The conversation emphasizes the importance of careful manipulation of complex functions in contour integration.
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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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You got as far as <br /> |\cot \pi z | = \left|\frac{e^{2i\pi z} + 1}{e^{2i\pi z} - 1} \right| = \left|<br /> \frac{e^{2i\pi x}e^{-2\pi y} + 1}{e^{2i\pi x}e^{-2\pi y} - 1}\right| 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 |\cot \pi z|^2, and at that point you can start trying to bound it on each side of the contour.

Or write <br /> |\cot \pi z| = \frac{|e^{2i\pi x} + e^{2\pi y}|}{|e^{2i\pi x} - e^{2\pi y}|} and then you can obtain an upper bound by maximizing the distance between e^{2i\pi x} and -e^{2\pi y} in the numerator and mimizing the distance between e^{2i\pi x} and e^{2\pi y} in the denominator.
 
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First, I tried to show that ##f_n## converges uniformly on ##[0,2\pi]##, which is true since ##f_n \rightarrow 0## for ##n \rightarrow \infty## and ##\sigma_n=\mathrm{sup}\left| \frac{\sin\left(\frac{n^2}{n+\frac 15}x\right)}{n^{x^2-3x+3}} \right| \leq \frac{1}{|n^{x^2-3x+3}|} \leq \frac{1}{n^{\frac 34}}\rightarrow 0##. I can't use neither Leibnitz's test nor Abel's test. For Dirichlet's test I would need to show, that ##\sin\left(\frac{n^2}{n+\frac 15}x \right)## has partialy bounded sums...