 Complex inverse cosine via the logarithm

Set $u=e^{iw}$ in $\cos w=z$. The resulting quadratic equation is $u^2-2zu+1=0$, with roots $z\pm i\sqrt{1-z^2}$. Their product is one, so taking a complex logarithm never encounters zero. Solving the exponential equation gives the displayed full inverse relation, with integer $n$. The principal complex logarithm and principal square root do not make both signs identical to the principal inverse cosine: at $z=0$ they yield $\pm \pi /2$, while principal $\arccos 0=\pi /2$. A single analytic branch requires compatible choices of sign and branches.