Suppose I wanna solve $e^{iz}=-1-\sqrt{2}$ now that is $$iz=\ln|-1-\sqrt{2}| +i(\operatorname{Arg}(-1-\sqrt{2})+2k\pi)$$ since $-1-\sqrt{2}$ is real $\operatorname{Arg}=0$,
so $z=-i\ln(1+\sqrt{2})+2k\pi$, $k$ is an integer. 

Am I wrong? it doesn't match the solution.