Really?

$\left( -1 \right) ={ e }^{ \left( 2k+1 \right) i\pi }, \quad k\in {Z}^{+} \Rightarrow \ln { \left( -1 \right) } =\ln { e^{ \left( 2k+1 \right) i\pi } } \Rightarrow \ln { \left( -1 \right) }=i\pi=0+i\pi$ $\text{Hence the real part is }\boxed0$

$\begin{aligned} z & = \ln(-1) \\ e^z & = -1 \\ & = - 1 + 0i \\ & = \cos \pi + i \sin \pi \\ \implies e^z & = e^{\pi i} \\ z & = \pi i \\ \implies \ln(-1) & = \color{#3D99F6}{0} + \pi i \end{aligned}$

Therefore, the real part of $\ln(-1)$ is $\boxed{0}$ .