# Stuck on proof using Cauchy's integral formula

I posted my attempted proof to this question here but I realized that I was wrong in taking the limit, and that the proof did not make sense. So I am still stuck on this problem

let $$f: \Omega \rightarrow \mathbb{C}$$ be analytic and $$z_0 \in \mathbb{C}$$.

Define $$g(z) = \begin{cases} \frac{f(z)-f(z_0)}{z- z_0} & z \not = z_0 \\ f'(z_0) & z = z_0 \end{cases}$$

now pick $$\varepsilon$$ small enough so that $$\overline{D(z_0, \varepsilon)} \subset \Omega$$

Show that whenever $$z \in D(z_0, \varepsilon)$$

$$\frac{g(z) - g(z_0)}{z-z_0} = \frac{1}{2\pi i}\int_{\partial D(z_0, \varepsilon)}\frac{f(\zeta)}{(\zeta-z)(\zeta-z_0)^2}d\zeta$$

So is Cauchy's integral formula still the right way to go? I end up getting that

$$\frac{g(z) - g(z_0)}{z-z_0} = \frac{f(z) - f(z_0)}{z-z_0} - \frac{1}{z-z_0}\int_{\partial D(z_0, \varepsilon)} \frac{f(\zeta)}{(\zeta-z_0)^2}d\zeta$$ and I am not sure how to proceed from here

## 2 Answers

Here is an approach that avoids the messy calculation. Fix $$z$$ and $$z_0$$ in $$D(z_0, \varepsilon)$$. Then, the residue theorem gives

$$\frac{1}{2\pi i}\int_{\partial D(z_0, \varepsilon)}\frac{f(\zeta)}{(\zeta-z)(\zeta-z_0)^2}d\zeta=\frac{f'(z_0)(z_0-z)-f(z_0)}{(z_0-z)^{2}}+\frac{f(z)}{(z-z_0)^{2}}=\frac{f(z)-f(z_0)}{(z-z_0)^{2}}-\frac{f'(z)}{z-z_0}.$$

On the other hand, by direct substitution,

$$\frac{g(z)-g(z_0)}{z-z_0}=\frac{f(z)-f(z_0)}{(z- z_0)^2}-\frac{f'(z_0)}{z-z_0}.$$

• wow thank you! I wouldn't have ever thought to have used the residue theorem – Richard Villalobos Dec 13 '18 at 14:39
• @RichardVillalobos I got to thinking more about the exercise because I wanted to work it as I think it was meant to be done. I have offered another answer below. – Matematleta Dec 14 '18 at 21:23

The answer above is slick, but if the exercise comes before the residue theorem, maybe we can brute force this. For convenience, write $$w$$ for $$\zeta$$. All integrals are over $$\partial D.$$ Fix $$z,z_0\in D.$$

It's easy to see that $$g$$ is analytic in $$\Omega$$, (it has a removable singularity at $$z=z_0.)$$

Then, we have

$$\tag1 g(z)=\frac{1}{2\pi i}\int \frac{g(w)dw}{w-z}$$

$$\tag2 f(z)=f(z_0)+(z-z_0)g(z).$$

All integrals of the form

$$\tag3 \int \frac{dw}{w-z}-\int \frac{dw}{w-z_0}$$

are equal to zero, because both $$z$$ and $$z_0$$ are contained in $$D$$.

Now, substitute $$(2)$$ into $$(1)$$ and use $$(3)$$ find that

$$g(z)=\frac{1}{2\pi i}\int \frac{f(w)dw}{(w-z_0)(w-z)}-\frac{1}{2\pi i}\int \frac{f(z_0)dw}{(w-z_0)(w-z)}=\frac{1}{2\pi i}\int \frac{f(w)dw}{(w-z_0)(w-z)}.$$

From here, using the fact that $$g(z_0)=f'(z_0)=\frac{1}{2\pi i}\int \frac{f(w)dw}{(w-z_0)^{2}}$$, it is a routine calculation:

$$\frac{g(z)-g(z_0)}{z-z_0}=\frac{1}{2\pi i(z-z_0)}\left [ \int \frac{f(w)dw}{(w-z_0)(w-z)}-\int \frac{f(w)dw}{(w-z_0)^{2}} \right ]=$$

$$\frac{1}{2\pi i(z-z_0)}\int\left [ \frac{f(w)(w-z_0)dw}{(w-z_0)^2(w-z)}-\int \frac{f(w)(w-z)dw}{(w-z_0)^{2}(w-z)} \right ]= \frac{1}{2\pi i}\int \frac{f(w)dw}{(w-z_0)^{2}(w-z)}$$