# Triangle Inequality Proof

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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The result you are trying to show is false. $b^2$ need not be real for arbitrary $b\in\mathbb{C}$. Take, for example, $b=1+i$, so $b^2=2i$. It is known that $\mathbb{C}$ is not an ordered field. Now, if you mean $|z^2+b^2|\geq|z|^2-|b|^2$, that is true. – Cameron Buie Apr 23 '12 at 19:01
Your attempt is damn wrong : $|1^2 + i^2| = 0$ but $|1|^2+|i|^2 = 2$... – Lierre Apr 23 '12 at 19:08
I've made the correction sorry guys. – Low Scores Apr 23 '12 at 19:54

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

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