I have the following cosmology exercise:
assume a flat Universe with a single component with an equation of state: \begin{equation} p = \omega \rho, \quad -1<\omega<1, \quad \omega = const. \end{equation} i) show that the angular diameter distance \begin{equation} d_A(z) = (1+z)^{-2} d_L(z) \end{equation} has always a turning point for $\omega >-1$.
My idea was to write the angular diameter distance as follow https://en.wikipedia.org/wiki/Angular_diameter_distance:
\begin{equation} d_A(z) = \frac{c}{H_0 q_0^2} \frac{zq_0 + (q_0-1)(\sqrt{2q_0+1} -1)}{(1+z)^2} \end{equation} and then replace for the deceleration parameter $q_0$ (choose for example $\Omega_m$ as the single component) (https://en.wikipedia.org/wiki/Deceleration_parameter)
\begin{eqnarray} q_0 &=& \frac{1}{2} \sum \Omega_i (1 + 3 \omega_i) \\ &=& \frac{1}{2} \Omega_m(1+3\omega_m) \end{eqnarray}.
Which would give:
\begin{equation} d_A(z) = \frac{c}{H_0 (\frac{1}{2} \Omega_m(1+3\omega_m))^2} \frac{z(\frac{1}{2} \Omega_m(1+3\omega_m)) + (\frac{1}{2} \Omega_m(1+3\omega_m)-1)(\sqrt{2 \cdot \frac{1}{2} \Omega_m(1+3\omega_m)+1} -1)}{(1+z)^2} \end{equation} where $\Omega_m = 1$ because it's a single component universe (is that correct?):
\begin{equation} d_A(z) = \frac{c}{H_0 (\frac{1}{2} 1(1+3\omega_m))^2} \frac{z(\frac{1}{2} 1(1+3\omega_m)) + (\frac{1}{2} 1(1+3\omega_m)-1)(\sqrt{2 \cdot \frac{1}{2} 1(1+3\omega_m)+1} -1)}{(1+z)^2} \end{equation}
and then derive this according to $z$ and replace the derivative with zero to find the solution. However I am not sure this is the right idea, The derivative will get really ugly, and I don't understand how the $\omega>-1$ conditions is relevant... And I don't know how to use the equation of state for ma calculation...Could anyone help me ?