= Solution
Similarly,
$$
1-z=e^{i\theta/2}\left(e^{-i\theta/2}-e^{i\theta/2}\right)
=-2i\sin\frac\theta2\,e^{i\theta/2}
=2\sin\frac\theta2\,e^{i(\theta/2-\pi/2)}.
$$
Thus, for $0<\theta<\pi$,
$$
\boxed{|1-z|=2\sin\frac\theta2,
\qquad \arg(1-z)=\frac\theta2-\frac\pi2}.
$$
At $\theta=0$, $1-z=0$, whose <complex argument> is undefined. Geometrically, $|1-z|$ is the chord joining the two unit-circle points $z$ and $1$. The isosceles triangle with vertex angle $\theta$ has half-chord $\sin(\theta/2)$, and the chord direction is perpendicular to the radius through its midpoint, giving the stated argument.
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