Introductio in analysin infinitorum

EX CIRCULO ORTIS. 105 posito $y$ loco $1+x$. Nunc igitur, posito loco $y$, partim $\cos.z + \sqrt{-1}. \sin.z$ partim $\cos.z - \sqrt{-1}. \sin.z$, prodibit $1 = \frac{1 + \frac{1}{i} l(\cos.z + \sqrt{-1}. \sin.z) + 1 + \frac{1}{i} l(\cos.z - \sqrt{-1}. \sin.z)}{2} = 1$, ob Logarithmos evanescentes, ita ut hinc nil sequatur. Altera vero aequatio pro Sinu suppeditat: $\frac{z}{i} = \frac{\frac{1}{i} l(\cos.z + \sqrt{-1}. \sin.z) - \frac{1}{i} l(\cos.z - \sqrt{-1}. \sin.z)}{2}$ ideoque $z = \frac{1}{2\sqrt{-1}} l \frac{\cos.z + \sqrt{-1}. \sin.z}{\cos.z - \sqrt{-1}. \sin.z}$, unde patet quemadmodum Logarithmi imaginarii ad Arcus circulares revocentur. 140. Cum fit $\frac{\sin.z}{\cos.z} = \operatorname{tang}.z$, Arcus $z [TRUNCATED]