Summation Formula (Complex Analysis)/Lemma

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Theorem

Let $N \in \N$ be an arbitrary natural number.

Let $C_N$ be the square embedded in the complex plane with vertices $\paren {N + \dfrac 1 2} \paren {\pm 1 \pm i}$.


Then there exists a constant real number $A$ independent of $N$ such that:

$\cmod {\map \cot {\pi z} } < A$

for all $z \in C_N$.


Proof

Let $z = x + iy$ for real $x, y$.


Case 1: $y > \frac 1 2$

We have:

\(\ds \cmod {\map \cot {\pi z} }\) \(=\) \(\ds \cmod {\frac {e^{i \pi z} + e^{-i \pi z} } {e^{i \pi z} - e^{-i \pi z} } }\) Euler's Cotangent Identity
\(\ds \) \(=\) \(\ds \cmod {\frac {e^{i \pi x - \pi y} + e^{-i \pi x + \pi y} } {e^{i \pi x - \pi y} - e^{-i \pi x + \pi y} } }\) as $z = x + iy$
\(\ds \) \(\le\) \(\ds \frac {\cmod {e^{i \pi x - \pi y} } + \cmod {e^{-\pi i x + \pi y} } } {\cmod {e^{-i \pi x + \pi y} } - \cmod {e^{i \pi x - \pi y} } }\) Triangle Inequality
\(\ds \) \(=\) \(\ds \frac {e^{-\pi y} + e^{\pi y} } {e^{\pi y} - e^{-\pi y} }\)
\(\ds \) \(=\) \(\ds \frac {e^{-2 \pi y} + 1} {1 - e^{-2 \pi y} }\) multiplying top and bottom by $\dfrac {e^{-\pi y} } {e^{-\pi y} }$
\(\ds \) \(\le\) \(\ds \frac {1 + e^{-\pi} } {1 - e^{-\pi} }\)
\(\ds \) \(=\) \(\ds A_1\)

$\Box$


Case 2: $y < -\frac 1 2$

Similarly:

\(\ds \cmod {\map \cot {\pi z} }\) \(=\) \(\ds \frac {\cmod {e^{i \pi x - \pi y} } + \cmod {e^{-\pi i x + \pi y} } } {\cmod {e^{-i \pi x + \pi y} } - \cmod {e^{i \pi x - \pi y} } }\)
\(\ds \) \(=\) \(\ds \frac {e^{-\pi y} + e^{\pi y} } {e^{-\pi y} - e^{\pi y} }\)
\(\ds \) \(=\) \(\ds \frac {1 + e^{2 \pi y} } {1 - e^{2 \pi y} }\) multiplying top and bottom by $\dfrac {e^{\pi y} } {e^{\pi y} }$
\(\ds \) \(\le\) \(\ds \frac {1 + e^{-\pi} } {1 - e^{-\pi} }\)
\(\ds \) \(=\) \(\ds A_1\)

$\Box$


Case 3: $-\frac 1 2 \le y \le \frac 1 2$

First consider $z = N + \frac 1 2 + iy$.

Then:

\(\ds \cmod {\map \cot {\pi z} }\) \(=\) \(\ds \cmod {\cot \pi \paren {N + \frac 1 2 + i y} }\)
\(\ds \) \(=\) \(\ds \cmod {\map \cot {\frac \pi 2 + i \pi y} }\) Cotangent Function is Periodic on Reals
\(\ds \) \(=\) \(\ds \cmod {\tanh \pi y}\) Cotangent of Complement equals Tangent, Hyperbolic Tangent in terms of Tangent
\(\ds \) \(=\) \(\ds \cmod {\tanh \frac \pi 2}\)
\(\ds \) \(=\) \(\ds A_2\)

Similarly in the case of $z = -N - \frac 1 2 + iy$, we have:

\(\ds \cmod {\map \cot {\pi z} }\) \(=\) \(\ds \cmod {\cot \pi \paren {-N - \frac 1 2 + i y} }\)
\(\ds \) \(=\) \(\ds \cmod {\map \cot {-\frac 1 2 + i y} }\) Cotangent Function is Periodic on Reals
\(\ds \) \(=\) \(\ds \cmod {\tanh \pi y}\) Cotangent of Complement equals Tangent, Hyperbolic Tangent in terms of Tangent
\(\ds \) \(=\) \(\ds \cmod {\tanh \frac \pi 2}\)
\(\ds \) \(=\) \(\ds A_2\)

$\Box$


Picking $A = \map \max {A_1, A_2}$ gives the desired bound.

$\blacksquare$


Sources

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