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I've been working through Marsden's Complex Analysis book, and I've come to a question where I'm not quite sure if I've got it. I would love some help.

question: Let f be analytic on a region, A, and let g be a closed curve in A. For any $z_0$ in A not on g, show that:

$\int_{g}\frac{f^{'}(a)}{{a - z_0}}da$ = $\int_{g}\frac{f(a)}{{(a - z_0)}^2}da$

and the followup question: how can you generalize this result?

my answer:

I thought I could use Cauchy's integral formula for derivatives, with the fact that given the function is analytic, that all of the derivatives of f exist on A and all are analytic.

So using the formula

$f^{k}(z_0) * I(g, z_0)$ = $\frac{k!}{2\pi i}$ $\int_g$$\frac{f(a)}{(a-z)^{k+1}}da;$ k = 1, 2, 3, ...

I just substituted in $f$ = $f^{'}$ to get the left hand side, since $f^{'}$ is analytic as well, and used k = 0; then I used $f = f$ and k = 1 to get the right hand side. This makes the equality hold.

I was not sure exactly how to generalize this, as I was kind of just figuring out a way to make the equality hold. This makes me think I'm doing it wrong.

many thanks.

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  • $\begingroup$ A generalization can be made with the same two steps but using the $i+1$ derivative and $k=0$ in the first step and using $k=i+1$ in the second. $\endgroup$ – Rafa Budría Mar 1 '17 at 9:42
  • $\begingroup$ clever, thank you very much. does the rest look correct? In particular, cauchy's integral formula works since f is analytic on A. is that enough to show that f's derivatives are all analytic as well? or do we need something else, like f is homotopic to a point in A? My book uses that language $\endgroup$ – mac5 Mar 1 '17 at 18:30
  • $\begingroup$ edit: here i means imaginary number. is that what you meant as well in your comment? or did you mean an index? $\endgroup$ – mac5 Mar 1 '17 at 18:47
  • $\begingroup$ Analytic in $A$ is enough $\endgroup$ – Rafa Budría Mar 1 '17 at 18:49
  • $\begingroup$ also, using the formula you said above, wouldnt it be off by a k! ? since the left side has k = 0 but the right side has k = i + 1 $\endgroup$ – mac5 Mar 1 '17 at 18:52
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Let's re-start. The formula is about derivatives. You didn't write it correctly (but your idea is good) The lhs has the k-th derivative

$$f^{(k)}(z_0) * I(g, z_0)=\frac{k!}{2\pi i}\oint_g\frac{f(a)}{(a-z)^{k+1}}da\tag 1$$

being $f^{(0)}(z_0)=f(z_0)$. Now, define $h(z)=f'(z)$, then:

$$h(z_0) * I(g, z_0)=\frac{1}{2\pi i}\oint_g\frac{h(a)}{a-z}da\tag 2$$

$$f'(z_0) * I(g, z_0)=\frac{1}{2\pi i}\oint_g\frac{f'(a)}{a-z}da$$

Taking $(1)$ With $k=1$:

$$f'(z_0) * I(g, z_0)=\frac{1}{2\pi i}\oint_g\frac{f(a)}{(a-z)^2}da$$

And you are done:

$$\oint_g\frac{f'(a)}{a-z}da=\oint_g\frac{f(a)}{(a-z)^2}da$$

The generalization. Let be $h(z)=f^{(j)}(z)$ and using $(2)$

$$f^{(j)}(z_0) * I(g, z_0)=\frac{1}{2\pi i}\oint_g\frac{f^{(j)}(a)}{a-z}da$$

Now, by $(1)$ with $k=j$

$$f^{(j)}(z_0) * I(g, z_0)=\frac{j!}{2\pi i}\oint_g\frac{f(a)}{(a-z)^{j+1}}da$$

Leading to:

$$\oint_g\frac{f^{(j)}(a)}{a-z}da=j!\oint_g\frac{f(a)}{(a-z)^{j+1}}da$$

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