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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.

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    $\begingroup$ $z_1$ and $z_2$ are the roots of the equation $z^2 \color{red}{-2\alpha} z+{\beta}^2=0$, not $z^2+2\alpha z+{\beta}^2=0$. I've fixed it. $\endgroup$
    – luxerhia
    Dec 9 '20 at 13:42
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    $\begingroup$ Fix $z_1+z_2=2\alpha$ too. $\endgroup$ Dec 9 '20 at 13:49
  • $\begingroup$ done! Thanks for the fixes $\endgroup$ Dec 9 '20 at 13:52
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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 $$ |\alpha -w | + |\alpha+w| = |\alpha + \beta| + |\alpha-\beta| \, . $$

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