Proving that $\left|\frac{1-e^{iz}}{z^2}\right| \le \frac{2}{|z|^2}$?

Question

I am wondering if I'm missing some sort of triangle inequality trick to show this, or if I need more context. Can this be solved purely algebraically?

Thanks.

Edit: I've tried this, but I'm not sure I'm using the triangle inequality properly.

$\left|\frac{1-e^{iz}}{z^2}\right| \le \frac{1}{|z|^2} + \left|\frac{-e^{iz}}{z^2}\right| = \frac{1}{|z|^2} + \frac{|-e^{iz}|}{|z^2|} = \frac{1}{|z|^2} + \frac{1}{|z^2|} = \frac{2}{|z|^2}$

Answer 1

The usual triangle inequality is enough, as the domain of this inequality is $|z|\ne 0$, it amounts to proving $$\bigl|1-\mathrm e^{iz}\bigr|\le 2. $$ Now, say $z=x+iy$, we have $$\bigl|1-\mathrm e^{iz}\bigr|\le 1+ \bigl|\mathrm e^{iz}\bigr|=1+\bigl|\mathrm e^{ix}\bigr|\bigl|\mathrm e^{-y}\bigr|=1+\bigl|\mathrm e^{-y}\bigr|,$$ so it is true if $y\ge 0$.

If $y<0$, it is easy to see that for each value of the argument, there exists a value of $z$ for which the inequality is false.