Bishop derivation completing the square in variational inference I don't understand the derivation on page 467. Bishop says:
Given the optimal factor $q_1^*(z_1)$
\begin{equation}
ln~q_1(z_1) = -\frac{1}{2} z_1^2 \Lambda_{11} + z_1 \mu_1 \Lambda_{11} - z_1 \Lambda_{12}( \mathbb{E}[z_2] - \mu_2) + cst
\end{equation}
using the technique completing the square, we can identify the mean and precision of this gaussian, giving:
\begin{equation}
    q^*(z_1) = \mathcal{N}(z_1 \mid m_1, \Lambda_{11}^{-1})
\end{equation}
where
\begin{equation}
    m_1 = \mu_1 - \Lambda_{11}^{-1} \Lambda_{12}( \mathbb{E}[z_2] - \mu_2)
\end{equation}
I don't really understand this method completing the square in this situation and how he got $m_1$

$$
\begin{align}
ln~q_1(z_1) &= -\frac{1}{2} z_1^2 \Lambda_{11} + z_1 \mu_1 \Lambda_{11} - z_1 \Lambda_{12}( \mathbb{E}[z_2] - \mu_2) + const. \\
    &= -\frac{1}{2} \Lambda_{11} \left(z_1^2 - 2\Lambda_{11}^{-1}z_1 \mu_1\Lambda_{11} + 2\Lambda_{11}^{-1}z_1\Lambda_{12}( \mathbb{E}[z_2] - \mu_2)\right) + const. \\
    &= -\frac{1}{2} \Lambda_{11} \left(z_1^2 - 2z_1(\mu_1 - \Lambda_{11}^{-1}\Lambda_{12}( \mathbb{E}[z_2] - \mu_2))\right) + const. \\
\end{align}
$$
now, let's say
$$
m_1 = \mu_1 - \Lambda_{11}^{-1}\Lambda_{12}( \mathbb{E}[z_2] - \mu_2)
$$
then we can rewrite previous equation in terms of $m_1$
$$
\begin{align}
ln~q_1(z_1) &= -\frac{1}{2} \Lambda_{11} \left(z_1^2 - 2z_1m_1\right) + const. \\
&= -\frac{1}{2} \Lambda_{11} \left(z_1^2 - 2z_1m_1 + m_1^2 - m_1^2\right) + const. \\
&= -\frac{1}{2} \Lambda_{11} \left(z_1^2 - 2z_1m_1 + m_1^2\right) + \frac{1}{2} \Lambda_{11}m_1^2 + const. \\
&= -\frac{1}{2} \Lambda_{11} (z_1 - m_1)^2 + \frac{1}{2} \Lambda_{11}m_1^2 + const. \\
&= -\frac{1}{2} \Lambda_{11} (z_1 - m_1)^2 + const. \\
\end{align}
$$
Now you can see that $const.$ will be just a part of the normalisation constant of the gaussian distribution, $m_1$ is a mean of the gaussian and $\Lambda_{11}$ is a precision
