Prove :
$\dfrac13\geq|1/(z^2-1)|$ for every $z$ on the circle $C(0,2)$
by the reverse triangle inequality: $|z_1-z_2|^2 \geq |z_1|^2-|z_2|^2$
Prove $\dfrac13\geq$ $|1/(z^2-1)|$ for every z on the circle $C(0,2)$
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complex-analysis
2 Answers
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If $|z| = 2$, then $|z|^2 = 4$ so $|z|^2-1 = 3$ so $|z^2-1| \ge |z|^2-1 =3 $, so, taking reciprocals $\dfrac1{|z^2-1|} \le \dfrac13 $.
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Reverse triangle inequality is $|z_1-z_2|\geq|z_1|-|z_2|.$
For $z \in C(0,2)$ one has $|z|=2,$ thus $$|z^2-1| \geq |z^2|-|1|=|z|^2-1=3.$$