= Solution
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> is
$$
d_A=\frac1{H_0u}\int_1^u v^{-\alpha}\,dv
=\begin{cases}
\dfrac{u^{-\alpha}-u^{-1}}{H_0(1-\alpha)},&\alpha\ne1,\\[4pt]
\dfrac{\log u}{H_0u},&\alpha=1.
\end{cases}
$$
It 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>:
$$
\boxed{z_m=\alpha^{1/(\alpha-1)}-1,\quad \alpha=\frac32(1+w),\qquad z_m=e-1\text{ when }\alpha=1.}
$$
For $\alpha\ne1$ 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 $\log(1+z_m)=\log\alpha/(\alpha-1)$:
$$
\frac{d\log(1+z_m)}{d\alpha}
=\frac{1-1/\alpha-\log\alpha}{(\alpha-1)^2}<0,
$$
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 $\sqrt3-1$ as $w\uparrow1$. 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.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2004/iii/paper-67-angular-distance-turning-point.png]
{title=Redshift of the angular diameter distance maximum versus the constant equation-of-state parameter}
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