Math Revision: Eigenvalues of a Two-by-Two-Matrix

$\displaystyle \hat{A}\underline{\phi} = \lambda \underline{\phi},\quad \hat{A} = \left(\begin{array}{cc} a & b \\ b^* & c \end{array} \right).$

(83)

$\displaystyle (\hat{A} - \lambda \hat{1}) \underline{\phi}$	$\displaystyle =$	$\displaystyle \underline{0}\leadsto \det (\hat{A} - \lambda \hat{1}) = 0$
$\displaystyle \leadsto \left\vert\begin{array}{cc} a-1 & b \\ b^* & c-1 \end{array} \right\vert$	$\displaystyle =$	$\displaystyle 0 \leadsto (a-\lambda)(c-\lambda)-\vert b\vert^2=0$
$\displaystyle \leadsto$		$\displaystyle \lambda^2 - (a+c)\lambda + ac - \vert b\vert^2 = 0$
$\displaystyle \lambda_{\pm}$	$\displaystyle =$	$\displaystyle \frac{a+c}{2}\pm \sqrt{\frac{(a-c)^2}{4}+\vert b\vert^2}.$	(84)