Complex numbers focus on the main Argument

Solution 1:

$\DeclareMathOperator{Arg}{Arg}$You are right that this can be done by considering three cases, but it is not necessary to compute $\Arg(1-z)$ explicitly.

$\Arg(z) + \Arg(1-z)$ is an argument of $z(1-z)$. It order to show that it is equal to $\Arg(z(1-z))$, it suffices to show that $\Arg(z) + \Arg(1-z)$ is in the range $(-\pi, \pi)$ for all $z$ in the given domain.

Case 1: $z$ is in the upper half-plane. Then $1-z$ is in the lower half-plane and $$ \Arg(z) \in (0, \pi) \, , \, \Arg(1-z) \in (-\pi, 0) \\ \implies \Arg(z) + \Arg(1-z) \in (-\pi, \pi) \, . $$

Case 2: $z$ is in the lower half-plane. This works similarly as in case 1.

Case 3: $z$ is a real number in the interval $(0, 1)$. Then $$ \Arg(z) + \Arg(1-z) = 0 + 0 = 0\, . $$