Firstly, for the first quadrant, we have $Re(z) > 0$ and $Im(z) > 0$.
Secondly, we have $z = \frac{1}{12}log(2) + (\frac{\pi}{8} + \frac{2\pi }{6} n)i$, $n \in \mathbb{Z}$ by part (a).
By combining the previous two conclusions we have, $Re(z) = \frac{1}{12}log(2) > 0$ since $log(2) > 0$. Also $Im(z) = (\frac{\pi}{8} + \frac{2\pi }{6} n) > 0$ when $n \ge 0$.
Therefore, as long as we have a non-negative $n$, our $z$ is in the first quadrant of complex plane.
