# 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

$$\displaystyle f'(a)=\frac{1}{2\pi i}\oint_C \frac{f(z)}{(z-a)^2}dz$$

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 $$f(a)=\frac{1}{2πi}\oint_C \frac{f(z)}{z-a}\,dz$$ .....(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?

$$\displaystyle f'(a)=\lim_{\Delta z\to0} \frac{f(a+\Delta z)-f(a)}{\Delta z}$$.

Now using (1) and on simplification gives

$$f'(a)=\lim_{\Delta z\to 0}\frac{1}{2\pi i}\oint_C\frac{f(z)}{(z-a-\Delta z)(z-a)}\,dz$$ which gives the required value as $$\Delta z→0$$. Is there any estimation really needed?

• How do you know that you can interchange the limit with the integral? And you're missing the $f(z)$ in the numerator of the integral expression. I've edited to correct the omission. Commented Mar 29, 2017 at 17:38
• Thanks for edit and pointing out the flaw in my work. Hope you suggest a way from this point to prove the result. Thanks again. Commented Mar 29, 2017 at 18:00
• Though it may not be with the weakest assumptions possible, I have seen a proof that I like using differential geometry and Stoke's theorem. I think you can find it towards the beginning of Chern's very nice book "Complex Manifolds without Potential Theory". Commented Mar 29, 2017 at 18:29

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}