Introductio in analysin infinitorum

98 DE QUANTITATIBUS TRANSCENDENT. LIB. I. Cum autem sit $cof. y. cof. z - fin. y. fin. z = cof. (y+z)$ & $cof. y. fin. z + fin. y. cof. z = sin. (y+z)$ erit hoc productum $(cof. y + \sqrt{-1}. fin. y) (cof. z + \sqrt{-1}. fin. z) = cof. (y+z) + \sqrt{-1}. fin. (y+z)$ & simili modo $(cof. y - \sqrt{-1}. fin. y) (cof. z - \sqrt{-1}. fin. z) = cof. (y+z) - \sqrt{-1}. fin. (y+z)$ item $(cof. x \pm \sqrt{-1}. fin. x) (cof. y \pm \sqrt{-1}. fin. y) (cof. z \pm \sqrt{-1}. fin. z) = cof. (x+y+z) \pm \sqrt{-1}. fin. (x+y+z)$. 133. Hinc itaque sequitur fore $(cof. z \pm \sqrt{-1}. fin. z)^2 = cof. 2z \pm \sqrt{-1}. fin. 2z$, & $(cof. z \pm \sqrt{-1}. fin. z)^3 = cof. 3z \pm \sqrt{-1}. fin. 3z$. ideoque generaliter erit $(cof. z \pm \sqrt{-1}. fin. z)^n = cof. nz \pm \sqrt{-1}. fin. nz:$ Unde, ob signorum ambiguitatem, erit $cof. nz = \frac{(cof. z + \sqrt{-1}. fin. z)^n + (cof. z - \sqrt{-1}. fin. z)^n}{2}$ & $fin. nz = \frac{(cof. z + \sqrt{-1}. fin. z)^n - (cof. z - \sqrt{-1}. fin. z)^n}{2\sqrt{-1}}$ Evolutis ergo binomiis hisce erit per Series: $cof. nz = (cof. z)^n - \frac{n(n-1)}{1.2} (cof. z)^{n-2} (fin. z)^2 + \frac{n(n-1)(n-2)(n-3)}{1.2.3.4} (cof. z)^{n-4} (fin. z)^4 - \frac{n(n-1)(n-2)(n-3)(n-4)(n-5)}{1.2.3.4.5.6} (cof. z)^{n-6} (fin. z)^6 + \&c., \&$ $fin. nz = \frac{n}{1} (cof. z)^{n-1} fin. z - \frac{n(n-1)(n-2)}{1.2.3} (cof. z)^{n-3} (fin. z)^3 + \frac{n(n-1)(n-2)(n-3)(n-4)}{1.2.3.4.5} (cof. z)^{n-5} (fin. z)^5 - \&c.$ 134.