 Solution

Suppose $|g|$ has a local maximum at $z_0$. If $g(z_0)=0$, the nonnegativity of the modulus forces $g=0$ throughout a neighbourhood, and the identity theorem for holomorphic functions makes it zero on that connected component.

If $g(z_0)\ne 0$, take $c=2i/g(z_0)$. Then $cg(z_0)=2i$, and continuity gives a disk $D$ around $z_0$ on which $\Im (cg)>1$. This lies within the given half-plane for a branch of the holomorphic logarithm. On $D$ defineComposition preserves holomorphy, and the real logarithm is strictly increasing, so $\Re f$ has a local maximum at $z_0$. Part (i) forces $f$ to be constant on the disk. Exponentiating gives $cg=e^f$, hence $g$ is constant there and, by the identity theorem for holomorphic functions, on the whole connected component. ThereforeThis proves the maximum modulus principle by the requested logarithm argument. Multiplication by $c$ is what puts the image in the supplied logarithm branch; no global logarithm of $g$ is assumed.