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Backgourd:

The complex function $f(z)$ is analytic on a closed circle $\overline D(z_0,r)$. Prove that: $$f'(z_0)=\frac{1}{2\pi\text{i}}\int_{\partial D(z_0,r)}\frac{f(z)}{(z-z_0)^{2}}\mathrm{d}z$$

We know that $f^{}(z_0)=\frac{1}{2\pi\text{i}}\int_{\partial D(z_0,r)}\frac{f(z)}{(z-z_0)^{}}\mathrm{d}z$, if we can "somehow" change the order of derivative: $$\tag{1}\frac{\mathrm{d}f}{\mathrm{d}z}_{|z=z_0}=\frac{1}{2\pi\text{i}}\frac{\mathrm{d}}{\mathrm{d}z}_{|z=z_0}\int_{\partial D(z_0,r)}\frac{f(\zeta)}{(\zeta-z_0)^{}}\mathrm{d}\zeta"="\frac{1}{2\pi\text{i}}\int_{\partial D(z_0,r)}\frac{f(\zeta)}{(\zeta-z_0)^{2}}\mathrm{d}\zeta,$$ then we are done. To prove the order can be changed, let $z_0+re^{\text{i}t}$ be the parametric representation of the circle $D(z_0,r)$. We also know: $$2\pi f(z_0)=\int_{0}^{2\pi}f(z_0+re^{\text{i}t})\mathrm{d}t\Rightarrow f'(z_0)=\frac{\mathrm{d}}{\mathrm{d}z}_{|z=z_0}\int_0^{2\pi}f(z_0+re^{\text{i}t})\mathrm{d}t$$ If the order can be changed, then $$\tag{2} \frac{\mathrm{d}}{\mathrm{d}z}_{|z=z_0}\int_0^{2\pi}f(z_0+re^{\text{i}t})\mathrm{d}t=\int_0^{2\pi}\frac{\partial f(z_0+re^{\text{i}t})}{\partial z}_{|z=z_0}\mathrm{d}t$$ But it seems to be miss something in $(2)$. What should I do if I want to solve my question by proving the odrer can be changed?

And here my textbook prove a lemma without showing the details of how to use the lemma to prove. enter image description here

What should I do to use the lemma to prove it?

Gary
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Qo Ao
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  • There are several instances of holomorphic differentiation under the integral being discussed on the site, e.g my answer here. Alternatively you can expand the denominator as a power series and exchange the series with an integral (using uniform convergence of the series) then easily read off the various derivatives (which I wrote in detail here); in fact this approach also shows that every holomorphic function on an annulus has a Laurent expansion. – peek-a-boo Aug 22 '24 at 04:06
  • For differentiation under the integral, even if you do not know about the dominated convergence theorem used in the link, you could try (with more effort) to prove a more elementary variant under stronger hypotheses like uniform convergence of the difference quotients; I leave this to you. – peek-a-boo Aug 22 '24 at 04:12
  • hmmm ok actually your very first step itself needs to be improved. While what you wrote using Cauchy’s formula for $f(z_0)$ is true, in order to differentiate, you need a formula for all $z$ near $z_0$, namely (by deforming the contour slightly) $f(z)=\frac{1}{2\pi i}\int_{|\zeta-z_0|=r}\frac{f(\zeta)}{\zeta-z},d\zeta$. Notice the denominator has $z$ not $z_0$, while the contour is a circle centered at $z_0$. Anyway, your approach (2) doesn’t really seem to work (atleast I don’t see right away how to make it work) because the derivative will land on the $f$ inside the integral. – peek-a-boo Aug 22 '24 at 04:18
  • @peek-a-boo Thanks a lot!Your comment reminds me of $\frac{f(z+h)-f(z)}{h}=\frac{1}{2\pi i}\int_{|w-z|=R}f(w)\frac{dw}{(w-z-h)(w-z)}$,and let$h\to 0$.But can i use the dominated convergence theorem here?Since $|f(w)|\le M$ is bounded and let $|h|<\frac{R}{2} $,thus$|w-z-h|\ge ||w-z|-|h||=R-|h|\ge \frac R2$,then$|f(w)\frac{1}{(w-z-h)(w-z)}|\le \frac{2M}{R^2}$ – Qo Ao Aug 22 '24 at 05:23
  • Yup, and the constant function $2M/R^2$ of course has finite arclength along the circle, so you can apply DCT now. – peek-a-boo Aug 22 '24 at 06:45

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