Showing convergence with complex numbers I would like to show that if $|a-b|<\delta$ then $|e^{a}-e^b|<\epsilon$. Where $a$ is complex and $b$ is real.
In essence if the difference between a and b is small then the difference between e^a and e^b is small
I was hoping to use The Mean value theorem and say the following:
Take $f(y)=e^y$ then $|f(a)-f(b)|=|a-b||f'(u)|$ and so as $|a-b|<\delta$ then $|f(a)-f(b)|<\epsilon$
unfortunately the mean value theorem follows a different structur when dealing with complex values (according to https://en.wikipedia.org/wiki/Mean_value_theorem)
I have a feeling that my statement is not necessarily true because if i write the $|e^a-e^b|$ as a sum I get 
$$
\left|\sum_{n=0}^{\infty} \frac{(a^n-b^n)}{n!}\right| \leq \sum_{n=0}^{\infty} \frac{(|a^n-b^n|)}{n!}
$$ 
and just because $|a-b|<\epsilon$ we don't necessarily  have $|a^n-b^n|<\epsilon$ as $n$ goes to $\infty$
Any help would be much appreciated 

edit:
in original i wrote "I would like to show that if $|a-b|<\epsilon$ then $|e^{a}-e^b|<\epsilon$. Where $a$ is complex and $b$ is real."
 A: Another approach:
$$
\begin{align}
\left|e^a-e^b\right|
&=\left|b-a\right|\left|\int_0^1e^{bt+a(1-t)}\,\mathrm{d}t\,\right|\\
&\le\left|b-a\right|\int_0^1e^{\mathrm{Re}(bt+a(1-t))}\,\mathrm{d}t\\[6pt]
&\le\left|b-a\right|e^{\max(\mathrm{Re}(a),\mathrm{Re}(b))}
\end{align}
$$
A: Here is a brute force approach.  Let $z=a-b$ with $\text{Re}(z)=x$ and $\text{Im}(z)=y$.  

NOTE:
In the ensuing development, we will use the inequalities 
$$\begin{align}
(1-e^{-|x|})^2&\le x^2 \tag 1\\\\
\sin^2(y)&\le y^2 \tag 2
\end{align}$$
which can be easily established a number of ways.  For example, $(1)$ can be shown from the inequality established in This Answer using the limit definition of the exponential function and Bernoulli's Inequality.   And $(2)$ can be established from elementary geometry.  Alternatively, both can be shown using the mean value theorem.

We begin with the term $|e^a-e^b|$ and write
$$\begin{align}
|e^a-e^b|&=|e^b||e^z-1|\\\\
&=|e^b||(e^x\cos(y)-1)+ie^x\sin(y)|\\\\
&=|e^b|\sqrt{e^{2x}-2e^x\cos(y)+1}\\\\
&=|e^b|\sqrt{(e^x-1)^2+2e^x(1-\cos(y))}\\\\
&=|e^b|\sqrt{(e^x-1)^2+4e^x\sin^2(y/2)} \tag 3\\\\
&=|e^b|e^x\sqrt{(1-e^{-x})^2+4e^{-x}\sin^2(y/2)} 
\tag 4
\end{align}$$ 
For $x\ge 0$, we have using $(4)$
$$\begin{align}
|e^b|e^x\sqrt{(1-e^{-x})^2+4e^{-x}\sin^2(y/2)}&\le |e^b|\,e^x\,\sqrt{x^2+y^2}\\\\
&=e^{x+\text{Re}(b)}|z|\\\\
&=e^{\text{Re}(a)}|a-b|
\end{align}$$
For $x<0$, we have using $(3)$
$$\begin{align}
|e^b|\sqrt{(e^x-1)^2+4e^x\sin^2(y/2)}&=|e^b|\sqrt{(1-e^{-|x|})^2+4e^{-|x|}\sin^2(y/2)}\\\\
&\le |e^b|\sqrt{x^2+y^2}\\\\
&= e^{\text{Re}(b)}|z|\\\\
&=e^{\text{Re}(b)}|a-b|
\end{align}$$
And now one can easily construct a $\delta-\epsilon$ proof.
