Simple proof of Cauchy Integral formula for derivatives While going through the different texts on Complex analysis, I encountered several ways to prove the formula
$f'(a)=\frac{1}{2πi}\int_C \frac{f(z)}{(z-a)^2}dz$
where $f(z)$ is analytic within and on simple closed curve $C$ and $a$ is any point inside $C$.
In every proof of this in different texts, I found one thing common that is obviously the Cauchy first formula $f(a)=\frac{1}{2πi}\int_C \frac{f(z)}{z-a}\,dz$ .....(1)
and an estimation for the absolute value of an integral which must go to zero (estimation theorem) in order to prove it.
I wonder whether or not my following proof is correct?
$f'(a)=\lim_{\Delta z→0} \frac{f(a+\Delta z)-f(a)}{\Delta z}$.
Now using (1) and on simplification gives 
$f'(a)=\lim_{\Delta z\to 0}\frac{1}{2\pi i}\int_C\frac{f(z)}{(z-a-\Delta z)(z-a)}\,dz$ which gives the required value as $\Delta z→0$. Is there any estimation really needed?
 A: We need to prove that differentiation under the integral is permissible.  To do so, we will estimate the difference between the difference quotient, $\frac{f(a+\Delta z)-f(a)}{\Delta z}$,and the result we obtain from differentiating under the integral, $\frac{1}{2\pi i}\oint_C \frac{f(z)}{(z-a)^2}\,dz$.
We begin with Cauchy's Integral formula.  If $f$ is analytic, then 
$$f(a)=\frac{1}{2\pi i}\oint_C \frac{f(z)}{z-a}\,dz \tag 1$$
We will show now that $f'(a)=\frac1{2\pi i}\oint_C\frac{f(z)}{(z-a)^2}\,dz$.
From $(1)$, we find that  
$$\begin{align}
\left|\underbrace{\frac{f(a+\Delta z)-f(a)}{\Delta z}}_{\to f'(a)\,\,\text{if the limit exists}}-\underbrace{\frac{1}{2\pi i}\oint_C \frac{f(z)}{(z-a)^2}\,dz}_{\text{Differentiating under the integral}}\right|&=\left|\frac{\Delta z}{(2\pi i)}\oint_C \frac{f(z)}{(z-a-\Delta z)(z-a)^2}\,dz\right|\\\\
&\le \frac{|\Delta z|}{2\pi}\oint_C \frac{|f(z)|}{|z-a|^2|z-a-\Delta z|}\,|dz| 
\end{align}$$
Since $f$ is continuous, it is bounded by say $B$.  Hence, $|f(z)|\le B$ for $z\in C$.  
Denote $L$ to be the length of the contour $C$ and denote $r$ to be the shortest distance between $a$ and any point on $C$.  
We take $|\Delta z|<r$.  Using the estimates $|z-a|\ge r$ and from the triangle inequality $|z-a-\Delta z|\ge ||z-a|-|\Delta z||\ge r-|\Delta z|$ reveals
$$\begin{align}
\left|\frac{f(a+\Delta z)-f(a)}{\Delta z}-\frac{1}{2\pi i}\oint_C \frac{f(z)}{(z-a)^2}\,dz\right|&\le \frac{|\Delta z|BL}{2\pi r^2(r-|\Delta z|)}\to 0\,\,\text{as}\,\,\Delta z\to 0
\end{align}$$
Therefore, we have proven that 
$$f'(a)=\frac1{2\pi i}\oint_C\frac{f(z)}{(z-a)^2}\,dz$$
and therefore
$$\begin{align}
f'(a)&=\lim_{\Delta z\to 0}\frac{f(a+\Delta z)-f(a)}{\Delta z}\\\\
&=\lim_{\Delta z\to 0}\frac{1}{2\pi i}\oint_C\frac{f(z)}{(z-a)(z-a-\Delta z)}\,dz\\\\
&=\frac{1}{2\pi i}\oint_C \lim_{\Delta z\to 0}\left(\frac{f(z)}{(z-a)(z-a-\Delta z)}\right)\,dz\\\\
&=\frac{1}{2\pi i}\oint_C\frac{f(z)}{(z-a)^2}\,dz
\end{align}$$
