Page:Joseph Louis de Lagrange - Œuvres, Tome 3.djvu/127

Or, comme

$\sin t={\frac {e^{t{\sqrt {-1}}}-e^{-t{\sqrt {-1}}}}{2{\sqrt {-1}}}},$

on aura

${\frac {1}{1+nz\sin t}}={\frac {1}{1+{\dfrac {nz}{2{\sqrt {-1}}}}\left(e^{t{\sqrt {-1}}}-e^{-t{\sqrt {-1}}}\right)}}\,;$

cette fraction peut se partager en celles-ci

$\alpha +\beta \left({\frac {1}{p+qe^{t{\sqrt {-1}}}}}+{\frac {1}{p-qe^{-t{\sqrt {-1}}}}}\right),$

en faisant

$p^{2}-q^{2}=1,\quad pq={\frac {nz}{2{\sqrt {-1}}}},\quad \alpha \left(p^{2}-q^{2}\right)+2p\beta =1,\quad \alpha pq+\beta q=0\,;$

d’où l’on tire

$\alpha =-{\frac {1}{p^{2}+q^{2}}},\quad \beta ={\frac {p}{p^{2}+q^{2}}},$

${\begin{aligned}p+q{\sqrt {-1}}=&{\sqrt {1+nz}},\\p-q{\sqrt {-1}}=&{\sqrt {1-nz}},\end{aligned}}$

et par conséquent

${\begin{aligned}p=&{\frac {{\sqrt {1+nz}}+{\sqrt {1-nz}}}{2}},\\q=&{\frac {{\sqrt {1+nz}}-{\sqrt {1-nz}}}{2{\sqrt {-1}}}}.\end{aligned}}$

Maintenant on aura

${\frac {1}{p+qe^{t{\sqrt {-1}}}}}={\frac {1}{p}}-{\frac {qe^{t{\sqrt {-1}}}}{p^{2}}}+{\frac {q^{2}e^{2t{\sqrt {-1}}}}{p^{3}}}-{\frac {q^{3}e^{3t{\sqrt {-1}}}}{p^{4}}}+\ldots ,$

et de même

${\frac {1}{p-qe^{-t{\sqrt {-1}}}}}={\frac {1}{p}}+{\frac {qe^{-t{\sqrt {-1}}}}{p^{2}}}+{\frac {q^{2}e^{-2t{\sqrt {-1}}}}{p^{3}}}+{\frac {q^{3}e^{-3t{\sqrt {-1}}}}{p^{4}}}+\ldots \,;$

donc la fraction ${\frac {1}{1+nz\sin t}}$ deviendra

${\begin{aligned}{\frac {1}{p^{2}+q^{2}}}&\left(1-{\frac {q}{p}}e^{t{\sqrt {-1}}}\,\ \ +{\frac {q^{2}}{p^{2}}}e^{2t{\sqrt {-1}}}\ \ -{\frac {q^{3}}{p^{3}}}e^{3t{\sqrt {-1}}}\ \ +\ldots \right.\\&\ \,\quad +\left.{\frac {q}{p}}e^{-t{\sqrt {-1}}}+{\frac {q^{2}}{p^{2}}}e^{-2t{\sqrt {-1}}}+{\frac {q^{3}}{p^{3}}}e^{-3t{\sqrt {-1}}}+\ldots \right).\end{aligned}}$