ce qui donnera (Article VIII) la fraction suivante
![{\displaystyle {\frac {m}{z(1+nz\sin t)(1+n\cos t)}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ef05e4cde8ee5b34cced53f4ada618d862175762)
Or on a déjà trouvé (Article IX)
![{\displaystyle {\frac {1}{z(1+nz\sin t)}}={\frac {d\mathrm {Z} }{dz}}\left[{\frac {1}{\mathrm {Z} }}-{\frac {n}{\sqrt {-1}}}e^{t{\sqrt {-1}}}\ \,+{\frac {n^{2}\mathrm {Z} }{({\sqrt {-1}})^{2}}}e^{2t{\sqrt {-1}}}-\ldots \right.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d6a904d442fcb1167a4a5ac223f7c62a2a080507)
![{\displaystyle \left.+{\frac {n}{\sqrt {-1}}}e^{-t{\sqrt {-1}}}+{\frac {n^{2}\mathrm {Z} }{({\sqrt {-1}})^{2}}}e^{-2t{\sqrt {-1}}}+\ldots \right],}](https://wikimedia.org/api/rest_v1/media/math/render/svg/88fa6c5d783fc1d5881c4bd4651826d5318ce6c1)
et l’on trouvera, de la même manière, en mettant
à la place de ![{\displaystyle {\sqrt {1-n^{2}}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/642964d16cff6c9951a159cecc9f7a65803ac596)
![{\displaystyle {\frac {1}{z(1+n\cos t)}}={\frac {1}{m}}\left[1-{\frac {n}{1+m}}e^{t{\sqrt {-1}}}\ \,+{\frac {n^{2}}{(1+m)^{2}}}e^{2t{\sqrt {-1}}}-\ldots \right.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/21dff2a14f69c30d282f8eb4c9a0f7f90306339e)
![{\displaystyle \left.-{\frac {n}{1+m}}e^{-t{\sqrt {-1}}}+{\frac {n^{2}}{(1+m)^{2}}}e^{-2t{\sqrt {-1}}}-\ldots \right].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0020c08a429d0848f4ec1110e3a54c2cda024a68)
Donc, mnltipliant ces deux séries l’une par l’autre, on aura la valeur de la fraction
![{\displaystyle {\frac {m}{z(1+nz\sin t)(1+n\cos t)}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fb4eb2de8e78e6ff30e04aa020627f3511344dde)
laquelle sera exprimée de cette manière
![{\displaystyle {\begin{aligned}\mathrm {M} &+\,\ \mathrm {M} 'e^{t{\sqrt {-1}}}+\,\ \mathrm {M} ''e^{2t{\sqrt {-1}}}+\ \mathrm {M} '''e^{3t{\sqrt {-1}}}+\ldots \\&+\mathrm {N} 'e^{-t{\sqrt {-1}}}+\mathrm {N} ''e^{-2t{\sqrt {-1}}}+\mathrm {N} '''e^{-3t{\sqrt {-1}}}+\ldots ,\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2f61c5a1df8d6ec570db59d141c831dae401b4e3)
en supposant, pour abréger,
![{\displaystyle {\begin{aligned}\mathrm {M} \ =&{\frac {1}{\mathrm {Z} }}{\frac {d\mathrm {Z} }{dz}}\left[1+{\frac {n^{4}\mathrm {Z} ^{2}}{(1+m)^{2}({\sqrt {-1}})^{2}}}+{\frac {n^{6}\mathrm {Z} ^{4}}{(1+m)^{4}({\sqrt {-1}})^{4}}}+\ldots \right]\\\mathrm {M} '=&{\frac {n}{\mathrm {Z} }}{\frac {d\mathrm {Z} }{dz}}\left[-{\frac {1}{1+m}}-{\frac {\mathrm {Z} }{\sqrt {-1}}}-{\frac {n^{2}\mathrm {Z} ^{2}}{(1+m)({\sqrt {-1}})^{2}}}-{\frac {n^{4}\mathrm {Z} ^{3}}{(1+m)^{2}({\sqrt {-1}})^{3}}}-\ldots \right.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d2630ec1cb41341383ceada847d7cc535edb4577)
![{\displaystyle \left.+{\frac {n^{2}\mathrm {Z} }{(1+m)^{2}{\sqrt {-1}}}}-{\frac {n^{4}\mathrm {Z} ^{2}}{(1+m)^{3}({\sqrt {-1}})^{2}}}+{\frac {n^{6}\mathrm {Z} ^{3}}{(1+m)^{4}({\sqrt {-1}})^{3}}}-\ldots \right],}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3fe1c102cd02fd5fdf79e498824deb8e3ef24a18)