MHB How Does the Triangle Inequality Apply to Complex Numbers?

AI Thread Summary
The discussion focuses on proving the inequality 2|z||w| ≤ |z|^2 + |w|^2 for complex numbers z and w. Participants suggest using the non-negativity of (|z| - |w|)^2 to derive the result, expanding it to show that |z|^2 + |w|^2 - 2Re(zw*) is non-negative. An alternative approach involves expressing |z|^2 + |w|^2 in terms of their real and imaginary components, leading to the application of the arithmetic-geometric mean inequality. The conversation also touches on the derivation of the expression for (|z| - |w|)^2 and its relation to the real part of the product of z and the conjugate of w. The thread effectively explores different methods to establish the triangle inequality in the context of complex numbers.
Poirot1
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let z,w be complex numbers. Prove:

2|z||w| <_ |z|^2 + |w|^2
 
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Well, you could note that $(|z| - |w|)^2 \geq 0$ and work from there.
 
0<_ (|z|-|w|)^2=|z|^2+|w|^2-2Re(zw*).

Ok so I can say Re(zw*)<_|zw*|=|zw| but that doesn't quite get me there.
 
They are both real numbers, so when you expand $(|z| - |w|)^2$ you get $|z|^2 - 2|z| |w| + |w|^2 \geq 0$, hence the result.

There is the alternative of using that

$$|z|^2 + |w|^2 = \Re^2 (z) + \Im^2 (z) + \Re^2 (w) + \Im^2(w)$$

and

$$2|z||w| = 2 \sqrt{\Re^2 (z) + \Im^2 (z) } \sqrt{\Re^2 (w) + \Im^2 (w)} = 2 \sqrt{(\Re^2 (z) + \Im^2 (z))(\Re^2 (w) + \Im^2 (w))}.$$

Now, $a = \Re^2 (z) + \Im^2 (z)$ is a positive real number, same as $b = \Re^2 (w) + \Im^2 (w)$. This is the arithmetic geometric mean written in a different way:

$$\frac{a+b}{2} \geq \sqrt{ab} \implies \frac{\Re^2 (z) + \Im^2 (z) + \Re^2 (w) + \Im^2(w)}{2} \geq \sqrt{(\Re^2 (z) + \Im^2 (z))(\Re^2 (w) + \Im^2 (w))}.$$

How did you get that $(|z| - |w|)^2 = |z|^2 + |w|^2 - 2 \Re (z \bar{w})$?
 
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