= Solution
By <Euler's formula>, $z=e^{i\theta}$. Factoring out the half-angle gives
$$
1+z=e^{i\theta/2}\left(e^{-i\theta/2}+e^{i\theta/2}\right)
=2\cos\frac\theta2\,e^{i\theta/2}.
$$
Since $0\leq\theta<\pi$, the cosine is positive. Therefore
$$
\boxed{|1+z|=2\cos\frac\theta2,
\qquad \arg(1+z)=\frac\theta2}.
$$
On the <Argand diagram>, the points $0,1,z,1+z$ form a rhombus. Its diagonal from $0$ to $1+z$ bisects the angle between the unit vectors $1$ and $z$, so its direction is $\theta/2$; resolving either side along the diagonal gives its length $2\cos(\theta/2)$.
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