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I need this one result to do a problem correctly.

I want to show that for any $b \in \mathbb{C}$ and $z$ a complex variable:

$ |z^2 + b^2| \geq |z|^{2} - |b|^{2}$

My attempts have only led me to conclude that

$ |z^2 + b^2| > \frac{|z|^{2} + |b|^{2}}{2}$

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    I've made the correction sorr$y$ gu$y$s.2012-04-23

3 Answers 3

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$ |z|^2 = |z^2| = |z^2 + b^2 - b^2| \le |z^2 + b^2| + |b|^2 $ and your inequality follows. I suppose you are missing an $|\cdot|$ arround the $b$? For you cannot compare complex numbers well.

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The searched inequality is an instance of the Lipshitz inequality for the distance.

In the concrete case $||z_1|-|z_2||\leq|z_1-z_2|$ It is obtained by the triangle inequality $|z_1|\leq|z_1-z_2|+|z_2|$ and its symmetric under the exchange of $z_1$ and $z_2$

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we know ,(from vector algebra) that

$ |z^2 + b^2| \geq |z^{2}| - |b^{2}|$

and, that for any complex number $x$,

$|x^{2}| \geq |x|^{2}$

therefore, $ |z^2 + b^2| \geq |z^{2}| - |b^{2}| \geq |z|^{2} - |b|^{2}$

hence, $ |z^2 + b^2| \geq |z|^{2} - |b|^{2}$