For real $x$ let $r(x)$ be defined (implicitly) through
$$ e^{i\,x} = \left(1+i\,x\right)\,e^{\left(-(1/2)\,x^2+r(x)\right)}, $$ where $i$ is the complex unit. I need to prove that $$ \left|r(x)\right|\leq \left|x\right|^3,\quad \text{for} ~~\left|x\right|\leq 1. $$ Unfortunately I do not have any clue on this.
===Some computations====
Using expansion I get $$ r(x) = i\,x+\frac{1}{2}\,x^2-\ln(1+i\,x)=i\,\frac{x^3}{3}+\frac{x^4}{4}-i\,\frac{x^5}{5}-\frac{x^6}{6}+i\,\frac{x^7}{7}+\frac{x^8}{8}+... $$ that is $$ r(x)=\sum_{k=2}^{\infty}(-1)^k\frac{x^{2\,k}}{2\,k}+i\,\sum_{k=1}^{\infty}(-1)^{k+1}\frac{x^{2\,k+1}}{2\,k+1} = \frac{1}{2} \left(x^2-\log \left(x^2+1\right)\right)+i\,\left(x-\tan ^{-1}(x)\right) $$
so that
$$ |r(x)| = \sqrt{\frac{1}{4} \left(x^2-\log \left(x^2+1\right)\right)^2+\left(x-\tan ^{-1}(x)\right)^2} $$
I think that the inequality does not hold. Let's find $r(x)$ explicitly. Let $p(x)$ and $q(x)$ be real functions such that $r(x)=p(x)+iq(x)$. To find $p(x)$, we take modulus on given equation: $$ 1=\sqrt{1+x^2}e^{-\frac{1}{2}x^2+p(x)} $$ Then we get $$ p(x)=\frac{1}{2}x^2 - \frac{1}{2}\ln(1+x^2). $$ Substitute $p(x)$ to the original equation, then $$ e^{ix}=(1+ix)e^{-\frac{1}{2}\ln(1+x^2)+iq(x)}=\frac{1+ix}{\sqrt{1+x^2}}e^{iq(x)}=e^{i(\theta(x)+q(x))}, $$ where $\theta(x)$ is a real number such that $\cos\theta(x)=\dfrac{1}{\sqrt{1+x^2}}$ and $\sin\theta(x)=\dfrac{x}{\sqrt{1+x^2}}$. Thus $$ q(x)=x-\theta(x)+2n\pi $$ for some $n\in \mathbb{Z}$. Define a complex-valued function $f:\mathbb{R}\to\mathbb{C}$ by $$ f(x)=\frac{1}{2}x^2-\frac{1}{2}\ln(1+x^2)+i\left(x-\arccos\frac{1}{\sqrt{1+x^2}}+2018\pi\right), $$ then $f$ satisfies the original equation. However, $f(0)\ne 0$.