= Complex inverse cosine via the logarithm
{title2=$w=-i\operatorname{Log}(z\pm i\sqrt{1-z^2})+2\pi n$}
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 $\arccos0=\pi/2$. A single analytic branch requires compatible choices of sign and branches.
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