I'm trying to understand the way that the Higgs Mechanism is applied in the context of a U(1) symmetry breaking scenario, meaning that I have a Higgs complex field $\phi=e^{i\xi}\frac{\left(\rho+v\right)}{\sqrt{2}}
  $  and want to gauge out the $\phi_2$  field that is causing my off-diagonal term, in normal symmetry breaking. I present the following transformation rules that hold in order to preserve local gauge invariance in Spontaneous Symmetry breaking non 0 vev for $\phi_1$ :

$$
\begin{cases}
\phi\rightarrow e^{i\theta}\phi\\
A_{\mu}\rightarrow A_{\mu}-\frac{1}{q}\partial_{\mu}\theta\\
D_{\mu}=\partial_{\mu}+iqA_{\mu}
\end{cases}
$$

As I understand, the Higgs gauge fixing mechanism is used to specify the transformations that gauge $\phi_2$ away. The idea is that we want to look for the angle that gives us a Higgs field with one real degree of freedom, as in

$$
\phi\rightarrow e^{i\theta}\phi=e^{i\theta}e^{i\xi}\frac{\left(\rho+v\right)}{\sqrt{2}}=e^{i\theta^{'}}\frac{\left(\rho+v\right)}{\sqrt{2}}
$$

so

$$\begin{cases}
\phi\rightarrow e^{i\theta}\phi=e^{i\theta}e^{i\xi}\frac{\left(\rho+v\right)}{\sqrt{2}}=e^{i\theta^{'}}\frac{\left(\rho+v\right)}{\sqrt{2}}\\
A_{\mu}\rightarrow A_{\mu}-\frac{1}{q}\partial_{\mu}\theta^{'}
\end{cases}
 $$

where $\theta^{'}=\theta+\xi$. I omit some factors of $v$ on the exponential for the time being. That's what I see my books doing. For $\theta^{'}=0$ this becomes

$$
\begin{cases}
\phi\rightarrow\frac{\left(\rho+v\right)}{\sqrt{2}}\\
A_{\mu}\rightarrow A_{\mu}
\end{cases}
 $$
and the rest is the derived desired interactions and terms in general. I note that this does not preserve local gauge invariance because 
 
$$\begin{cases}
\phi\rightarrow e^{i\theta^{'}}\frac{\left(\rho+v\right)}{\sqrt{2}}\neq e^{i\theta^{'}}\phi\\
A_{\mu}\rightarrow A_{\mu}-\frac{1}{q}\partial_{\mu}\theta^{'}
\end{cases}
 $$

Am I looking at this the right way?