In Stein's Complex analysis book, example 2 of page 44 says
$f(z)$ = $\frac{(1-\exp(iz))}{z^2}$ = $\frac{-iz}{z^2}$ + $E(z)$
where $E(z)$ is bounded as z $\to$ 0.
Of course, z is complex variable.
I can not understand why $E(z)$ is bounded like that. Could you help me to understand this?
$$\frac{1 - e^{iz}}{z^2} = \frac{1 - (1 + iz + \frac{1}{z}(iz)^2 + \ldots)}{z^2} = -iz + \frac{1}{z^2}\sum_{k = 2}^{+\infty}\frac{(-iz)^k}{k!} $$
You can see pretty well that the term
$$\frac{1}{z^2}\sum_{k = 2}^{+\infty}\frac{(-iz)^k}{k!} = E(z)$$
Is bounded for $z\to 0$.
Expand the sum and watch.