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Im trying to solve a inequality from a book about Complex Analisys.

If $z\in\mathbb{C}$ then

$$ \bigg\vert\frac{1-e^{iz}}{z^2}\bigg\vert \leq \frac{2}{\big\vert z\big\vert^2} $$

im still tring to solve this.

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This statement is false. Take $z = -2i$. –  Qiaochu Yuan Sep 7 '12 at 2:26

2 Answers 2

up vote 1 down vote accepted

Well, you could observe that $$e^{-2y}-2e^{-y}\cos(x)-3$$ is a quadratic in $e^-y$. Make the substitution $u=e^{-y}$, and we must show that for all positive $u$ and all $x$, $$u^2-2\cos(x)u-3\leq 0.$$ However, this cannot possibly be true, since the left side tends to infinity as $u$ does. You must have made a mistake somewhere.

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this is very possible... –  Integral Sep 7 '12 at 2:03
    
there was an error indeed, but i think its not im my solution but in my problem, i saw wrong. Its not $2/|z^2|$, its $2/|z|^2$ and i think this will make difference. –  Integral Sep 7 '12 at 2:11
    
those two quantities are the same –  Zarrax Sep 7 '12 at 2:28

For $z\neq 0$, the inequality would be equivalent to $|1-e^{iz}|\leq 2$, hence we would have $|e^{i(x+iy)}|=e^{-y}\leq 3$ for all $y\in \Bbb R$, which can't be true. However, the initial inequality is true for $z\in \Bbb R$.

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