Find expression for $\arccos(z)$ with complex logarithm

Question

I want to find an expression for $\arccos(z)$ with complex logarithm.

I know that it can be expressed as the exponential whose exponent is proportional to a complex logarithm, but I do not know how to do it.

Answer 1

If $w=\arccos z,$ then $z=\cos w = \frac{e^{iw}+e^{-iw}}{2}.$ We solve this equation for $w$:

$$2z = e^{iw}+e^{-iw}$$

$$2ze^{iw} = e^{2iw}+1$$

$$(e^{iw})^2 -2ze^{iw} +1 = 0$$

$$e^{iw} = \frac{2z\pm \sqrt{(2z)^2-4}}{2}$$

$$e^{iw} = z\pm \sqrt{z^2-1}$$

$$iw = \ln(z\pm \sqrt{z^2-1})$$

$$w = \frac{\ln(z\pm \sqrt{z^2-1})}{i}.$$

Then think about the $\pm$ sign.