A complex numbers' modulus identity.

[Alone]

I am searching for a shortcut in the calculation of a proof.

The question is as follows:

2.12 Prove that:

$$|z_1|+|z_2| = |\frac{z_1+z_2}{2}-u|+|\frac{z_1+z_2}{2}+u|$$

where $z_1,z_2$ are two complex numbers and $u=\sqrt{z_1z_2}$.

I thought of showing that the squares of both sides of the above identity are equal, in which case since both sides of the above identity are nonnegative we will get the above identity.

The problem that it seems too tedious work, unless there's some trick to be used here?

 

[HallsofIvy]

You have [math]\left|\frac{z_1+ z_2}{2}- \sqrt{z_1z_2}\right|= \left|\frac{z_1- 2\sqrt{z_1z_2}+ z_2}{2}\right|[/math].

Noting the resemblance to [math]a^2+ 2ab+ b^2[/math] I would let [math]a= \sqrt{z_1}[/math] and [math]b= \sqrt{z_2}[/math]. Then [math]\left|\frac{z_1- 2\sqrt{z_1z_2}+ z_2}{2}\right|= \frac{(a- b)^2}{2}[/math]. Do the same thing for [math]\left|\frac{z_1+ z_2}{2}+ \sqrt{z_1z_2}\right|= \left|\frac{z_1+ 2\sqrt{z_1z_2}+ z_2}{2}\right|[/math]