= Solution
Normalize today's <scale factor> to $a_0=1$ and use $c=1$. A source of bolometric <luminosity> $L$ emits <photons> whose <energies> arrive smaller by $1+z$, and whose arrival intervals are longer by $1+z$. Their wavefront has physical area $4\pi a_0^2r_1^2$. The received <radiative flux> is therefore
$$
F=\frac{L}{4\pi a_0^2r_1^2(1+z)^2}.
$$
By the definition of <luminosity distance>, $F=L/(4\pi d_L^2)$, so $d_L=(1+z)a_0r_1=r_1/a_{\rm em}$. The final equality assumes the stated normalization of the <scale factor>; it is not a formula invariant under an arbitrary rescaling of $a_0$.
A radial <null geodesic> of the <FRW metric> obeys $dt/a=dr/\sqrt{1-Kr^2}$. Using $1+z=1/a$ and $dz/dt=-(1+z)H(z)$ gives the radial <comoving distance>
$$
\chi(z)=\int_{t_{\rm em}}^{t_0}\frac{dt}{a(t)}
=\int_0^z\frac{dz'}{H(z')}
=\int_0^{r_1}\frac{dr}{\sqrt{1-Kr^2}}.
$$
For $K>0$ this last integral is $\arcsin(\sqrt K r_1)/\sqrt K$, continued along the radial coordinate when necessary; for $K<0$ it is $\operatorname{arsinh}(\sqrt{|K|}r_1)/\sqrt{|K|}$. Inverting, with $\Omega_K=-K/H_0^2$, proves
$$
\boxed{d_L(z)=\frac{1+z}{H_0\sqrt{|\Omega_K|}}\,
S_K\!\left(H_0\sqrt{|\Omega_K|}\int_0^z\frac{dz'}{H(z')}\right).}
$$
Here $S_K$ is sine for positive <spatial curvature of an FLRW universe>, hyperbolic sine for negative <spatial curvature of an FLRW universe>, and its argument for zero <spatial curvature of an FLRW universe>. The flat case is the continuous limit $d_L=(1+z)\chi$, not a literal division by zero. The original PDF has $H_0\sqrt{|\Omega_K|}$ in the denominator and $H=\dot a/a$; the converted TeX loses these details.
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