Solution
= Solution
Set
$$
g(z)=\frac{f(z)-b}{a}.
$$
Then $g$ is entire and never takes a real value. The continuous function $\operatorname{Im}g$ is nowhere zero; because $\mathbb C$ is connected, it has one sign everywhere. Thus either $\operatorname{Im}g$ or $-\operatorname{Im}g$ is a positive <harmonic function>. Part (i) makes it constant. The <Cauchy-Riemann equations> then force the real part of $g$ to be constant as well. Hence $g$ and therefore $f$ are constant.