1
$\begingroup$

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 $z \in D(z_0, \varepsilon)$

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

My question is if my proof below, which makes use of the Cauchy integral formula is correct

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

now as epsilon gets arbitrarily small, $z \rightarrow z_0$ and so now making use of the Cauchy integral formula and the fact that $\lim_{z \rightarrow z_0} \frac{f(z_0)}{z-z_0} = 0$ we should have

$$\lim_{z \rightarrow z_0} \frac{f(z)-f(z_0)}{(z- z_0)^2} - \frac{f'(z_0)}{z-z_0} = \frac{1}{2 \pi i}\int_{D(z_0, \varepsilon)}\frac{f(\zeta)}{(\zeta-z)(\zeta-z_0)^2}d\zeta - 0$$

which gives us the result

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.