# Little Complex Inequality

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.

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

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.
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
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 $z\in \Bbb R$.