= Solution
For a single constant-$w$ <perfect fluid> in a flat expanding <FLRW metric> model, <cosmological energy conservation> gives $\rho=\rho_0a^{-3(1+w)}$ and $H(z)=H_0(1+z)^{3(1+w)/2}$. With $a_0=1$, the <luminosity distance> follows by integrating the radial <null geodesic>:
$$
D_L(z)=(1+z)\int_0^z\frac{dz'}{H(z')}.
$$
The <constant-equation-of-state luminosity distance> is consequently
$$
\boxed{D_L(z)=\begin{cases}
\displaystyle\frac{2(1+z)}{H_0(1+3w)}\left[1-(1+z)^{-(1+3w)/2}\right],&w\ne-1/3,\\[5pt]
\displaystyle\frac{1+z}{H_0}\log(1+z),&w=-1/3.
\end{cases}}
$$
This includes $w=-1$ regularly, giving $D_L=z(1+z)/H_0$. Restoring conventional units multiplies the distance formulas by $c$. Expanding about $z=0$ gives
$$
\boxed{D_L(z)=\frac1{H_0}\left[z+\frac{1-3w}{4}z^2+O(z^3)\right],\qquad
q_0=\frac{1+3w}{2}.}
$$
The second result also follows directly from the <Friedmann acceleration equation>. The expansion agrees with the general coefficient $(1-q_0)/2$ of $z^2$.
For <phantom energy>, $w<-1$, put $n=3(1+w)/2<0$. Integration of $\dot a=H_0a^{1-n}$ with $a(t_{\rm now})=1$ gives
$$
a(t)=\left[1+nH_0(t-t_{\rm now})\right]^{1/n}.
$$
Because $1/n<0$, the <scale factor> diverges at the <Big Rip> time
$$
\boxed{t_{\rm rip}-t_{\rm now}=\frac{2}{3|1+w|H_0}.}
$$
This is a property of the stipulated constant-$w$ fluid model; it is not realizable by the positive-density canonical scalar of the preceding parts.
\Image[/past-exam-of-the-general-relativity-and-cosmology-course-of-the-university-of-oxford/2013-constant-w-luminosity-distance.png]
{title=Exact flat single-fluid luminosity distances for several constant equations of state, including the logarithmic coasting case}
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