An approximation for $e^{ix}$ By the Taylor expansion of $\log(1+z)$ one can show that
$$e^{ix} = (1+ix)e^{-x^2/2+r(x)}$$
where $\lvert r(x)\rvert \leq \lvert x\rvert^3$ for all $x$ such that $\lvert x\rvert \leq 1/2$.
In a paper I saw the same claim except for the part where the author claims the the bound on $r(x)$ is true for all $x$. I have no clue how they arrived at this. Could this be an error? Or does someone have a proof for this?
 A: If we define \begin{align}
r(x)=ix+\frac{x^2}{2}-\log (1+ix)\tag{1}\end{align}
then $e^{ix} = (1+ix)e^{-x^2/2+r(x)}$.
Using the Taylor expansion of $\log(1+z)$ in $(1)$ we have \begin{align}
r(x)&= ix+\frac{x^2}{2}-\left(ix+\frac{x^2}{2}-i\frac{x^3}{3}-\frac{x^4}{4}-i\frac{x^5}{5}+\cdots\right)\\
&=i\frac{x^3}{3}+\frac{x^4}{4}+i\frac{x^5}{5}+\cdots.
\end{align}
Therefore we have
\begin{align}
|r(x)|&\le \frac{|x|^3}{3}+\frac{|x|^4}{4}+\frac{|x|^5}{5}+\cdots\le\frac{|x|^3}{3}+\frac{|x|^4}{3}+\frac{|x|^5}{3}+\cdots\\
&=\frac{|x|^3}{3}\left( 1+|x|+|x|^2+\cdots\right)=\frac{|x|^3}{3}\cdot\frac{1}{1-|x|}\\
&\le \frac{2\,|x|^3}{3}\le |x|^3
\end{align}
for $|x|\le \frac{1}{2}$.
EDIT:
If we want to argue it for all values of $x$, use inequalities \begin{align}
&|x^2-\log (1+x^2)|\le |x|^3,\tag{1}\\
&|x-\tan^{-1}x|\le \frac{|x|^3}{3}\tag{2}
\end{align}
instead of the Taylor expansion of $\log (1+z)$. 
Then \begin{align}
|r(x)|&=\left|ix+\frac{x^2}{2}-\log (1+ix)\right|\\
&=\left|\frac{\,1\,}{2}\left(x^2-\log (1+x^2)\right)+i(x-\tan^{-1} x)\right|\\
&\le \frac{\,1\,}{2}|x|^3+\frac{\,1\,}{3}|x|^3\le |x|^3.
\end{align}
