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$$\frac{f(a)}{(a-z)^{k+1}}da;$ k = 1, 2, 3, ...
I just substituted in $f$ = $f^{'}$ to get the left hand side, since $f^{'}$ is analytic as well, and used k = 0; then I used $f = f$ and k = 1 to get the right hand side. This makes the equality hold.
I was not sure exactly how to generalize this, as I was kind of just figuring out a way to make the equality hold. This makes me think I'm doing it wrong.
many thanks.
Let's re-start. The formula is about derivatives. You didn't write it correctly (but your idea is good) The lhs has the k-th derivative
$$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$$