In the solution of
$$ \int_0^{\infty}\frac{\sin(x)}{x} $$
The Conway book for complex analysis state the following:
Since $\displaystyle\frac{e^{iz} - 1}{z}$ has a removable singularity at $z =0$, then $\exists M > 0$ such that
$$ |\frac{e^{iz} - 1}{z}| \leq M, |z| \leq 1$$
Why this is true ?
I know that if $f(z)$ has a removable singularity at $z = 0$, then $\displaystyle\lim_{z\to 0} zf(z) = 0$, but how that guarantees that bound ?
The function $h:\mathbb C\to\mathbb R_+$ given by $$ h(x)=\begin{cases} \left|\displaystyle\frac{e^{iz} - 1}{z}\right|,&\ z\ne0\\ \ \\0,&\ z=0\end{cases} $$ is continuous. A continuous function maps compact sets to compact sets. So $h$ maps the closed unit disc $\overline{\mathbb D}$ to a compact subset of $[0,\infty)$. Compact sets in any metric space are bounded; so there exists an interval $[0,M]$ such that $h(\overline{\mathbb D})\subset [0,M]$. In other words, $$ h(z)\leq M, \ \ |z|\leq1. $$