# Summation Formula (Complex Analysis)/Lemma

## Theorem

Let $C_N$ be the square with vertices $\left({N + \frac 1 2}\right) \left({\pm 1 \pm i}\right)$ for $N \in \N$.

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

$\displaystyle \left\vert{\cot \left({\pi z}\right)}\right\vert < A$

for all $z$ on $C_N$.

## Proof

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

### Case 1: $y > \frac 1 2$

We have:

 $\displaystyle \left\vert{\cot \left({\pi z}\right)}\right\vert$ $=$ $\displaystyle \left\vert{\frac {e^{i \pi z} + e^{-i \pi z} } {e^{i \pi z} - e^{-i \pi z} } }\right\vert$ Cotangent Exponential Formulation $\displaystyle$ $=$ $\displaystyle \left\vert{\frac {e^{i \pi x - \pi y} + e^{-i \pi x + \pi y} } {e^{i \pi x - \pi y} - e^{-i \pi x + \pi y} } }\right\vert$ as $z = x + iy$ $\displaystyle$ $\le$ $\displaystyle \frac {\left\vert{e^{i \pi x - \pi y} }\right\vert + \left\vert{e^{-\pi i x + \pi y} }\right\vert} {\left\vert{e^{-i \pi x + \pi y} }\right\vert - \left\vert{e^{i \pi x - \pi y} }\right\vert}$ Triangle Inequality $\displaystyle$ $=$ $\displaystyle \frac {e^{-\pi y} + e^{\pi y} } {e^{\pi y} - e^{-\pi y} }$ $\displaystyle$ $=$ $\displaystyle \frac {e^{-2 \pi y} + 1} {1 - e^{-2 \pi y} }$ multiplying through $\dfrac {e^{-\pi y} } {e^{-\pi y} }$ $\displaystyle$ $\le$ $\displaystyle \frac {1 + e^{-\pi} } {1 - e^{-\pi} }$ $\displaystyle$ $=$ $\displaystyle A_1$

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

Similarly:

 $\displaystyle \left\vert{\cot \left({\pi z}\right)}\right\vert$ $=$ $\displaystyle \frac {\left\vert{e^{i \pi x - \pi y} }\right\vert + \left\vert{e^{-\pi i x + \pi y} }\right\vert} {\left\vert{e^{-i \pi x + \pi y} }\right\vert - \left\vert{e^{i \pi x - \pi y} }\right\vert}$ $\displaystyle$ $=$ $\displaystyle \frac {e^{-\pi y} + e^{\pi y} } {e^{-\pi y} - e^{\pi y} }$ $\displaystyle$ $=$ $\displaystyle \frac {1 + e^{2 \pi y} } {1 - e^{2 \pi y} }$ multiplying through $\dfrac {e^{\pi y} } {e^{\pi y} }$ $\displaystyle$ $\le$ $\displaystyle \frac {1 + e^{-\pi} } {1 - e^{-\pi} }$ $\displaystyle$ $=$ $\displaystyle A_1$

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

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

Then:

 $\displaystyle \left\vert{\cot \left({\pi z}\right)}\right\vert$ $=$ $\displaystyle \left\vert{\cot \pi \left({N + \frac 1 2 + iy}\right)}\right\vert$ $\displaystyle$ $=$ $\displaystyle \left\vert{\cot \left({\frac \pi 2 + i \pi y}\right)}\right\vert$ Cotangent Function is Periodic on Reals $\displaystyle$ $=$ $\displaystyle \left\vert{\tanh \pi y}\right\vert$ Cotangent of Complement equals Tangent, Hyperbolic Tangent in terms of Tangent $\displaystyle$ $=$ $\displaystyle \left\vert{\tanh \frac \pi 2}\right\vert$ $\displaystyle$ $=$ $\displaystyle A_2$

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

 $\displaystyle \left\vert{\cot \left({\pi z}\right)}\right\vert$ $=$ $\displaystyle \left\vert{\cot \pi \left({-N - \frac 1 2 + iy}\right)}\right\vert$ $\displaystyle$ $=$ $\displaystyle \left\vert{\cot \left({-\frac 1 2 + iy}\right)}\right\vert$ Cotangent Function is Periodic on Reals $\displaystyle$ $=$ $\displaystyle \left\vert{\tanh \pi y}\right\vert$ Cotangent of Complement equals Tangent, Hyperbolic Tangent in terms of Tangent $\displaystyle$ $=$ $\displaystyle \left\vert{\tanh \frac \pi 2}\right\vert$ $\displaystyle$ $=$ $\displaystyle A_2$

Picking $A = \max \left({A_1, A_2}\right)$ gives the desired bound.

$\blacksquare$