TITLE: Let $\alpha$ and $\beta$ be any two distinct complex numbers,then $|\alpha-\sqrt{\alpha^2-\beta^2}|+|\alpha+\sqrt{\alpha^2-\beta^2}|=$
QUESTION [7 upvotes]: Let $ \alpha $    and    $\beta  $ be any two distinct complex numbers,then $|\alpha-\sqrt{\alpha^2-\beta^2}|+|\alpha+\sqrt{\alpha^2-\beta^2}|=$
My Attempt
Let $z_1=\alpha-\sqrt{\alpha^2-\beta^2}$,
$z_2=\alpha+\sqrt{\alpha^2-\beta^2}$
$z_1$ and $z_2$ are the roots of the complex valued equation
$z^2-2\alpha z+{\beta}^2=0$
So
$z_1+z_2=2\alpha$
and
$z_1z_2={\beta}^2$
now since alpha and beta are complex we cannot assume the roots to be conjugate and hence
$\overline{z_1}=z_2$ does not hold and consequently I am unable to find the necessary equations to solve for the value of
$|z_1|+|z_2|$
Please help me out!
Thanks in advance.

REPLY [4 votes]: In order to simplify the calculation, and to avoid confusion with the two possible values of the complex square root, it is useful to define $w$ as a square root of $\alpha^2 - \beta^2$ (is does not matter which one), and then work only with the property that
$w^2 = \alpha^2 - \beta^2$.
Using the parallelogram law twice one gets
$$
\begin{align}
 \bigl( |\alpha -w | + |\alpha+w| \bigr)^2 &= |\alpha -w|^2 + |\alpha+w|^2 + 2|\alpha^2-w^2| \\
 &= 2 |\alpha|^2 + 2 |w|^2 + 2|\beta|^2 \\
 &= |\alpha  + \beta|^2 + 2 |\alpha^2 - \beta^2| + |\alpha-\beta|^2 \\
 &= \bigr(|\alpha  + \beta| + |\alpha-\beta| \bigr)^2
\end{align}
$$
and therefore
$$
|\alpha -w | + |\alpha+w| = |\alpha  + \beta| + |\alpha-\beta| \, .
$$