Let $\alpha$ and $\beta$ be any two distinct complex numbers,then $|\alpha-\sqrt{\alpha^2-\beta^2}|+|\alpha+\sqrt{\alpha^2-\beta^2}|=$

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

$$|z_1|+|z_2|$$

• $z_1$ 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. Dec 9, 2020 at 13:42
• Fix $z_1+z_2=2\alpha$ too.
– V.G
Dec 9, 2020 at 13:49
• done! Thanks for the fixes Dec 9, 2020 at 13:52

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 $$|\alpha -w | + |\alpha+w| = |\alpha + \beta| + |\alpha-\beta| \, .$$