MHB 1st Derivative of Cauchy Integral formula

[ognik]

Hi - I know the final result for the n'th derivative, I am looking though at getting an expression for the 1st derivative of f(z).

From $ f({z}_{0}) = \frac{1}{2\pi i} \oint_{c} \frac{f(z)}{z - {z}_{0}}dz $ we get

$ \frac{f({z}_{0} + \delta {z}_{0}) -{f({z}_{0}}) }{\delta {z}_{0}} =
\frac{1}{2\pi i \delta {z}_{0} } (\oint_{c} \frac{f(z)}{z - {z}_{0} - \delta {z}_{0} } dz - \oint_{c} \frac{f({z}_{0})}{z - {z}_{0}}dz ) $

Where does the 2nd integral on the right come from?

 

[Ackbach]

Hi - I know the final result for the n'th derivative, I am looking though at getting an expression for the 1st derivative of f(z).

From $ f({z}_{0}) = \frac{1}{2\pi i} \oint_{c} \frac{f(z)}{z - {z}_{0}}dz $ we get

$ \frac{f({z}_{0} + \delta {z}_{0}) -{f({z}_{0}}) }{\delta {z}_{0}} =
\frac{1}{2\pi i \delta {z}_{0} } (\oint_{c} \frac{f(z)}{z - {z}_{0} - \delta {z}_{0} } dz - \oint_{c} \frac{f({z}_{0})}{z - {z}_{0}}dz ) $

Where does the 2nd integral on the right come from?

I think the 2nd integral on the right should just be a copy of your original expression, right? That is, you should have
$$ \frac{f({z}_{0} + \delta {z}_{0}) -{f({z}_{0}}) }{\delta {z}_{0}} =
\frac{1}{2\pi i \delta {z}_{0} } \left(\oint_{c} \frac{f(z)}{z - {z}_{0} - \delta {z}_{0} } dz - \oint_{c} \frac{f({z})}{z - {z}_{0}}dz \right). $$

 

[ognik]

It's not quite the same, it has $ f({z}_{o}) $ on top instead of $ f({z}) $ and also the $\delta {z}_{o} $ in the divisor in front.

Saying that told me what I was missing (and also what you were saying), it is of course the $ - f({z}_{o}) $ part of the expression. Thanks Ackbach, got it now.