Really?-2

$\left( -1 \right) ={ e }^{ \left( 2k+1 \right) i\pi }, \quad k\in {Z}^{+}\Rightarrow \ln { \left( -1 \right) } =\ln { e^{ \left( 2k+1 \right) i\pi } }$ $\text{The first real value is at } k=0$ $\Rightarrow \ln { \left( -1 \right) }=i\pi \Rightarrow \boxed{A=\pi}$

$\begin{aligned} z & = \ln(-1) \\ e^z & = -1 \\ & = - 1 + 0i \\ & = \cos \pi + i \sin \pi \\ \implies e^z & = e^{\pi i} \\ z & = \ln(-1) = \pi i \\ \implies A & = \pi \approx \boxed{3.141} \end{aligned}$