How to calculate arccos of complex number I want to know what can I do to calculate the acos of a complex number. I looked at this answer on this site but that only work for imaginary numbers.
Given the complex number:
$$ \arccos(2 + 3i) $$
How can I calculate it?
Thanks in advance!

$z=\arccos(2+3i)$
$2+3i=\cos z=(e^{iz}+e^{-iz})/2$; $e^{iz}+e^{-iz}=4+6i$; $u^2-(4+6i)u+1=0$ where $u=e^{iz}$; $$u={4+6i\pm\sqrt{(4+6i)^2-4}\over2}={4+6i\pm\sqrt{-24+48i}\over2}=2+3i\pm\sqrt{-6+12i}$$ so we need to find $\sqrt{-6+12i}$.
Let $\sqrt{-6+12i}=r+si$. Then $-6+12i=r^2-s^2+2rsi$, so $$r^2-s^2=-6,\qquad 2rs=12$$ Then $$(r^2+s^2)^2=(r^2-s^2)^2+(2rs)^2=36+144=180$$ so $r^2+s^2=\sqrt{180}=6\sqrt5$. Then $2r^2=6\sqrt5-6$ and $2s^2=6\sqrt5+6$, so $$r=\sqrt{3\sqrt5-3},\qquad s=\sqrt{3\sqrt5+3}$$ So $e^{iz}=u=2+3i\pm(\sqrt{3\sqrt5-3}+\sqrt{3\sqrt5-3}i)$, and $z=-i\log u$.
But how do we find $\log u$, when $u$ is a complex number? Let's do it in general. Write $a+bi=\sqrt{a^2+b^2}e^{i\arctan(b/a)}$. Then $\log(a+bi)=(1/2)\log(a^2+b^2)+i\arctan(b/a)$.
