Let $a<b$ real numbers and let $H=\{z \in \mathbb{C}: \operatorname{Im}(z)>0\}$. Define $u: H \rightarrow \mathbb{R}$ such that $u(z)$ is represent the angle (in radians) between the vector $za$ and the vector $zb$. show that $u$ is harmonic.
What I understand that $$0<u<\pi$$ but it doesnt't seems to help me. so the only way I think is actually build the function with trigonometry relation and show it by direct calculation. This part is hard for me and I don't know how to actually do it.
Let $\operatorname{Log}(z)$ be the principal value of the logarithm in $H$, i.e. $$ \operatorname{Log} = \log |z| + i \operatorname{Arg}(z) \text{ with } 0 < \operatorname{Arg} < \pi \, . $$ $\operatorname{Log}$ is holomorphic in $H$, so that its imaginary part $\operatorname{Arg}$ is harmonic in $H$, and so is your function $$ u(z) = \operatorname{Arg}(z-b) - \operatorname{Arg}(z-a) \, . $$