Corollary 2

For every odd integer $n = 2m + 1$, and every complex number $z$ such that $nz$ is not an integral multiple of $\pi$, one has

(4)

Indeed, $\sin n \left( z + \frac{\pi}{2} \right) = \sin \left( nz + \frac{\pi}{2} + m\pi \right) = (-1)^m \cos nz$, whence, replacing $z$ by $z + \frac{\pi}{2}$ in (2),

(5)

and the formulae (2) and (5) give (4) on division term-by-term when $\sin nz \neq 0$.

In all that follows we shall always assume that $n = 2m + 1$ is an odd integer; the formula (4) can also be written

or, one has

for tan z finite; cot $nz$ is thus a rational function in $u = \tan z$, whose numerator is of degree $n - 1$, and whose denominator, of degree $n$, has the $n$ simple roots $\tan k\pi / n$; decomposing this into simple fractions yields

where

whence, on isolating the term in (6) corresponding to $k = 0$ and combining the terms corresponding to opposite values of $k$, and replacing $z$ by $z/n$,

valid for every complex number $z$ not an integral multiple of $\pi/2$. One can write this formula in the form

with $v_k(n, z) = 0$ for $k > m$ and

for $1 \leq k \leq m$. We shall see that for every $z$ contained in a compact subset K of $\mathbf{C}$, not containing any integral multiple of $\pi$, and for every sufficiently large odd $n$, the series with general term $v_k(n, z)$ is normally convergent. Indeed, as $n$ tends to $+\infty$, $\tan \frac{z}{n}$ tends to $\frac{z}{n}$ uniformly on K, so there exists a number $M > 0$ such that $\left| n \tan \frac{z}{n} \right| \leq M$ for every sufficiently large $m$ and every $z \in K$. On the other hand, for $0 \leq x \leq \pi/2$ one has $\sin x / x \geq 1 - \frac{x^2}{6} \geq \frac{1}{2}$, so for $1 \leq k \leq m$ one has $n \sin \frac{k\pi}{n} \geq k\pi/2$; consequently, when $m$ is sufficiently large, for every integer $k$ such that $k\pi /2 > M$ one has $|v_k(n, z)| \leq \frac{8M}{k^2 \pi^2 - 4M^2}$, which proves our assertion. For $k$ odd, $v_k(n, z)$ tends (uniformly on K) to $\frac{2z}{z^2 - k^2 \pi^2}$ as $n$ tends to $+\infty$. Consequently: