1st Extension of Cauchy Integral Formula

Question

I was trying to obtain first extension of Cauchy Integral Formula which is

$$ f'(z)= \frac{1}{2\pi i} \int_C \frac{f(s)}{(s-z)^2} ds$$

where $s$'s are points on $C$ contour and $z$ is any point in interior of $C$

I could write these ones only :

$$\frac{f(z+\Delta z)-f(z)}{\Delta z} = \frac{1}{2\pi i} \int_C \biggr( \frac{1}{s-z-\Delta z} - \frac{1}{s-z} \biggr) \frac{f(s)}{\Delta z} ds $$

$$= \frac{1}{2\pi i} \int_C \frac{f(s)}{(s-z-\Delta z)(s-z)} ds$$

I've seen that in Brown and Churchill's Complex Variables and Applications (8th Ed.) :

$$ \frac{f(z+\Delta z)-f(z)}{\Delta z} - \frac{1}{2\pi i} \int_C \frac{f(s)}{(s-z)^2} ds = \frac{1}{2\pi i} \int_C \frac{\Delta z.f(s)}{(s-z-\Delta z)(s-z)^2} ds$$

I know it's a basic question but I couldn't obtain in no way the last equation from my writtens. Could you please help me?

Thanks

Answer 1

What you did above is correct. Now substitute and you will get the desired expression.

\begin{align}&\frac{f(z+\Delta z)-f(z)}{\Delta z} - \frac{1}{2\pi i} \oint_C \frac{f(s)}{(s-z)^2} \,ds\\ &=\frac{1}{2\pi i} \oint_C \frac{f(s)}{(s-z-\Delta z)(s-z)}\, ds-\frac{1}{2\pi i} \oint_C \frac{f(s)}{(s-z)^2} \,ds\\ &= \frac{1}{2\pi i}\oint_C \frac{f(s)}{(s-z)}\left (\frac{1}{(s-z-\Delta z)}-\frac{1}{(s-z)}\right)\,ds\\ &=\frac{1}{2\pi i} \oint_C \frac{\Delta z.f(s)}{(s-z-\Delta z)(s-z)^2} \,ds \end{align}