I have understood up to the Cauchy Integral Formula, including it's derivation. I would like to understand the jump from the formula to the general expression, which utilizes the formula for derivatives. Book: Fundamentals of Complex Analysis with Applications to Engineering and Science by E.B. Saff and A.D. Snider.

Thusfar, $\zeta$ has represented the poles of a function. $\Gamma$ represents contours, $z$ inside of $\Gamma$.

The text gives Cauchy's Integral Formula:

$$f(z_0) = \frac{1}{2\pi i}\int\frac{f(z)}{z-z_0} dz$$

The text then says: "If in Cauchy's formula we replace $z$ by $\zeta$ and $z_0$ by $z$, then we obtain

$$f(z) = \frac{1}{2\pi i} \int \frac{f(\zeta)}{\zeta-z} d\zeta$$

The advantage of this representation is that it suggests a formula for the derivative $f'(z)$,obtained by formally differentiating with respect to z under the integral sign. thus we are led to suspect that

$$f'(z) = \frac{1}{2\pi i} \int\frac{f(\zeta)}{(\zeta-z)^2} d\zeta$$

In verifying this equation we shall actually establish a more general theorem..."

What I would like to understand is why these replacements happen, how this thought process came about, and how this necessarily differs from the original representation.

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    $\begingroup$ It's just changing letters to make de-emphasize the fixed point $z_0$. $\endgroup$ – T. Bongers Dec 22 '15 at 4:40
  • $\begingroup$ @user I figured as much, but wouldn't replacing $z_0$ with $\zeta$ make more sense, since $\zeta$ is a pole of a function (a point where the function isn't analytic), and leaving $z$ alone as it is actually $z(t)$ and parametrizes the contour? Essentially, in this case $\zeta$ has nothing to do with the zeros of a function? Thank you for responding. $\endgroup$ – amelia Dec 22 '15 at 4:49

In the first equation, $$f(z_0)=\frac 1{2\pi i}\int_\Gamma\frac{f(z)}{z-z_0}\,dz$$ we see that the value of the given integral is equal the value of the function $f$ at $z_0$. So, the value of $f$ at a single point $z_0$ is represented by the integral. Note here that $z_0$ is a constant.

Now, since $z$ represents complex values in the interior of $\Gamma$ we can view the Cauchy integral as representing the values for all points in the interior of $\Gamma$. Of course to do this, we must change the variable of integration. Hence $$f(z)=\frac1{2\pi i}\int_{\Gamma}\frac{f(\zeta)}{\zeta-z}\,d\zeta.$$

  • $\begingroup$ Thank you for your response. This makes sense to me, however I am still cloudy on the use of $\zeta$, as it has always been used in this text to represent the zero of denominators. $\endgroup$ – amelia Dec 22 '15 at 4:54
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    $\begingroup$ That's a matter of context. If it helps, change $\zeta$ to $w$. In fact, some texts write the Cauchy Integral Formula in this way. $\endgroup$ – Tim Raczkowski Dec 22 '15 at 4:56

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