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The cosmological perfect-fluid continuity equation gives $\rho \propto a^{-3(1+w)}$. In a flat universe the Friedmann equation then gives $H(z)=H_0(1+z{)}^{\alpha }$, where $\alpha =3(1+w)/2>0$. Writing $u=1+z$, the angular diameter distance obtained from the luminosity distance isIt is positive for $u>1$, starts at zero and tends to zero as $u\to \infty$ for every $\alpha >0$, including $\alpha <1$ where the radial comoving distance itself is unbounded. Differentiation gives the constant-equation-of-state angular diameter distance maximum:For $\alpha \ne 1$ the stationary equation is $u^{\alpha -1}=\alpha$, which has exactly one solution greater than one. For $\alpha =1$ the derivative has the sign of $1-\log u$. Hence this stationary point is always the unique maximum.

To determine the sketch's shape, differentiate $\log (1+z_m)=\log \alpha /(\alpha -1)$:with limiting value $-1/2$ at $\alpha =1$. The inequality follows because $\log \alpha -1+1/\alpha$ has its unique minimum zero at one. Thus $z_m$ decreases from infinity as $w\downarrow -1$ to $\sqrt{3}-1$ as $w\uparrow 1$. Useful points are $z_m=e-1$ at $w=-1/3$, $5/4$ at $w=0$, and $1$ at $w=1/3$. The endpoint $w=-1$ is excluded: there $d_A=z/[H_0(1+z)]$ increases to an asymptote and has no finite turning point.