A Taylor series expansion of $e^{ix}$ In Probability Theory by Athreya and Lahiri, they give a very elegant proof of Central Limit Theorem (The Lindeberg one) wherein they use a lemma: 
For $x \in \mathbb{R}$ and $r \geq 1$, ($i=\sqrt{-1}$ here)
$$\left|e^{ix} - \sum_{k=0}^{r-1} \frac{(ix)^k}{k!}\right| \leq \min\left\{\frac{|x|^r}{r!},\frac{2|x|^{r-1}}{(r-1)!}\right\}\quad (1)$$
In the proof of the lemma, they give the following expansion of $e^{ix}$ arguing that it follows from Taylor series. 
$$e^{ix} = \sum_{k=0}^{r-1} \left[\frac{(ix)^k}{k!}\right] + \frac{(ix)^r}{(r-1)!}\int_0^1(1-u)^re^{iux}du \quad (2)$$
By taking absolute values, they would get $(1)$. What I don't get is how did they arrive at $(2)$ in the first place. I have seen Taylor series (and Taylor's theorem) for real valued functions but not for complex valued functions. Moreover I don't recall an integral being used to approximate the remainder. 
Hence my question is twofold:
1) Does anyone know the equivalent of a Taylor Theorem for complex functions as above? 
2) Is there an alternative way to prove $(1)$ without resorting to Taylor Series? I tried to dominate the remainder terms by a geometric series but that failed.
Good references and proofs/hints for this are welcome. Let me know if you need more details.
Update1: In my second question, what I meant is "Is there an alternate way to prove (1)". You can use Taylor series and find a clever way to bound the higher order terms. As I mentioned, I used a geometric series bound but that didn't work. 
 A: By the fundamental theorem of calculus,
$$
e^{ix} = 1 + (ix) \int_{0}^{1} e^{iux}\, du,
$$
which is the case $r = 1$ of
$$
e^{ix} = \sum_{k=0}^{r-1} \left[\frac{(ix)^{k}}{k!}\right]
  + \frac{(ix)^{r}}{(r - 1)!} \int_{0}^{1}(1 - u)^{r-1} e^{iux}\, du.
\tag*{$P(r)$}
$$
(N.B. $(1 - u)^{r-1}$ in the integrand, not $(1 - u)^{r}$.)
Assume inductively that $P(r)$ is true for some $r \geq 1$. Integrating by parts with
\begin{align*}
 U &= e^{iux},         &  V &= -\frac{(1 - u)^{r}}{r}, \\
dU &= ix e^{iux}\, du, & dV &= (1 - u)^{r-1}\, du, \\
\end{align*}
gives
\begin{align*}
\frac{(ix)^{r}}{(r - 1)!} \int_{0}^{1}(1 - u)^{r} e^{iux}\, du
  &= \frac{(ix)^{r}}{(r - 1)!}
  \left[-e^{iux}\frac{(1 - u)^{r}}{r}\bigg|_{u=0}^{u=1}
  + \frac{(ix)}{r} \int_{0}^{1}(1 - u)^{r} e^{iux}\, du\right] \\
  &= \frac{(ix)^{r}}{r!}
  + \frac{(ix)^{r+1}}{r!} \int_{0}^{1}(1 - u)^{r} e^{iux}\, du.
\end{align*}
Substituting this into $P(r)$ gives $P(r + 1)$.
A: There are various forms for the remainder term of a finite Taylor expansion. One of them is
$$f(x)=\sum_{k=0}^nf^{(k)}(a){(x-a)^k\over k!} +\int_a^xf^{(n+1)}(t){(x-t)^n\over n!}\ dt\ ,\tag{1}$$
whereby $f$ is assumed sufficiently differentiable in the neighborhood of $x=a$. This is of course valid as well if $f$ is complex-valued.
For the proof of $(1)$ begin with
$$f(x)=f(a)+\int_a^xf'(t)\cdot 1\ dt$$
and set up an induction, using repeated partial integration.
