MHB A complex numbers' modulus identity.

AI Thread Summary
The discussion focuses on proving the identity involving the modulus of complex numbers, specifically that |z_1| + |z_2| equals the sum of two expressions involving the average of z_1 and z_2 and the square root of their product. The approach suggested involves squaring both sides of the identity to simplify the proof, leveraging the non-negativity of the terms. A connection to the algebraic identity a^2 + 2ab + b^2 is noted, where a and b are the square roots of z_1 and z_2, respectively. The discussion highlights the potential complexity of the proof and seeks a more efficient method to demonstrate the identity. Ultimately, the conversation emphasizes the need for a clever shortcut in the proof process.
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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?
 
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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]
 
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