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Inequality of complex numbers

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Poirot

Banned
Feb 15, 2012
250
let z,w be complex numbers. Prove:

2|z||w| <_ |z|^2 + |w|^2
 

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
Well, you could note that $(|z| - |w|)^2 \geq 0$ and work from there.
 
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Poirot

Banned
Feb 15, 2012
250
0<_ (|z|-|w|)^2=|z|^2+|w|^2-2Re(zw*).

Ok so I can say Re(zw*)<_|zw*|=|zw| but that dosnt quite get me there.
 

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
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 $(|z| - |w|)^2 = |z|^2 + |w|^2 - 2 \Re (z \bar{w})$?