Introductio in analysin infinitorum

104 DE QUANTATIBUS TRANSCENDENT. LIB. I. $\frac{(1 + \frac{v\sqrt{-1}}{i})^i + (1 - \frac{v\sqrt{-1}}{i})^i}{2}$; atque $fin. v = \frac{(1 + \frac{v\sqrt{-1}}{i})^i - (1 - \frac{v\sqrt{-1}}{i})^i}{2\sqrt{-1}}$. In Capite autem præcedente vidimus esse $(1 + \frac{z}{i})^i = e^z$, denotante $e$ basin Logarithmorum hyperbolicorum : scripto ergo pro $z$ partim $+v\sqrt{-1}$ partim $-v\sqrt{-1}$ erit $cof. v = \frac{e^{+v\sqrt{-1}} + e^{-v\sqrt{-1}}}{2}$ & $fin. v = \frac{e^{+v\sqrt{-1}} - e^{-v\sqrt{-1}}}{2\sqrt{-1}}$. Ex quibus intelligitur quomodo quantitates exponentiales imaginariæ ad Sinus & Cosinus Arcuum realium reducantur. Erit vero $e^{+v\sqrt{-1}} = cof. v + \sqrt{-1}.fin. v$ & $e^{-v\sqrt{-1}} = cof. v - \sqrt{-1}.fin. v$. 139. Sit jam in iisdem form [TRUNCATED]