$\frac{1 - e^{iz}}{z^2} = \frac{-iz}{z^2} + E(z)$ where $E(z)$ is bounded as $z \rightarrow 0?$ Was reading some notes and it states that $f(z) = \frac{1 - e^{iz}}{z^2}$ can be written as $f(x) = \frac{-iz}{z^2} + E(z)$ where $E(z)$ is bounded as $z \rightarrow 0.$ I don't exactly see why. Help is appreciated. 

Taylors series of $e^{iz}$ about $z=0$ is $$1+iz+\frac{(iz)^2}{2!}+\frac{(iz)^3}{3!}+\cdots$$ so $$\frac{1-e^{iz}}{z^2}=\frac{1}{z^2}\left(1-1-iz-\frac{(iz)^2}{2!}-\frac{(iz)^3}{3!}-\cdots\right)=\frac{-iz}{z^2}+E(z)$$ with $E(z)=-\frac{1}{z^2}\left(\frac{(iz)^2}{2!}+\frac{(iz)^3}{3!}+\cdots\right)$.
