= Solution
For a finite separation use the Cartesian distance, rather than substituting finite coordinate differences into the local <line element>. Expanding the two horizontal components gives
$$
(\rho\cos\phi-\rho'\cos\phi')^2+(\rho\sin\phi-\rho'\sin\phi')^2
=\rho^2+\rho'^2-2\rho\rho'\cos(\phi-\phi').
$$
The vertical contribution is $(z-z')^2$, independently. Thus
$$
\boxed{h=\rho^2+\rho'^2-2\rho\rho'\cos(\phi-\phi'),\qquad j=(z-z')^2,}
$$
and
$$
\boxed{R=\sqrt{\rho^2+\rho'^2-2\rho\rho'\cos(\phi-\phi')+(z-z')^2}.}
$$
Equivalently, the horizontal contribution is $(\rho-\rho')^2+4\rho\rho'\sin^2[(\phi-\phi')/2]$, which also makes its nonnegativity explicit.
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