Complex contour integral proof

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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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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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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