I got 2 complex roots as well. It looks like there's a slight typo, should be $e^z = \dfrac{-1 \pm i\sqrt{3}}{2}$, so we get the same answer as the solution: $log\left(\dfrac{-1 \pm i\sqrt{3}}{2}\right) = \left(\pm\dfrac{2}{3} + 2n\right)i\pi,$ for $n\in\mathbb{Z}$.
