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I am trying to calculate the following integral $\int_\gamma \frac{1}{(z-w)^2} dw$ with $\gamma$ being a circle around some point $a$, e.g. $\gamma(t) := a+re^{it}$, $t \in [0, 2\pi]$.

Instead of using the Cauchy-Integralformula I need to compute this integral via finding the potency series for $\frac{1}{(z-w)^2}$

First of all, I don't see, how developing the series, using the formula $c_n = \frac{1}{2\pi i} \int_{|z-z_0|} \frac{f(z)}{(z-z_0)^{(n+1)}} dz$ for the coefficients would help me to solve the integral.

Furthermore I can't find another way of doing this, without using the Cauchy-Formula for the coefficients, like it is possible for finding a series for $\frac{1}{w-z}$.

I hope some of you can help me.

Thanks!

EDIT: Some addenta (Thanks to @José Carlos Santos)

Due to: $\frac{1}{w-z} = -\frac{1}{z}\frac{1}{1-w/z} = -\frac{1}{z}(1+\frac{w}{z}+\frac{w^2}{z^2}+...)$ is $\frac{1}{(w-z)^2}=\frac{d}{dw}(\frac{1}{w-z}) = \frac{1}{z^2}+\frac{2w}{z^3}+\frac{3w^2}{z^4}+...$

The geometrical sum converges uniformly, so it is allowed to switch summation and differentiation.

Now we have:

\begin{align} \int_{\gamma} \left( \frac{1}{z^2}+\frac{2w}{z^3}+\frac{3w^2}{z^4}+... \right) dw &= \frac{1}{z^2} \left( \int_{0}^{2\pi} r~i~e^{it} dt~+~ \frac{2}{z} \int_0^{2\pi}(a+r~e^{it})~r~i~e^{it} dt+ ... \right) \\ &= 0 + 0 + ... = 0 \end{align}

So the integral equals zero, being independent of the location of the pole? This doesn't seem to be consistent with Cauchy's integral-formula, or Cauchy's residue formula respectively, doesn't it?

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Since$$\frac1{w-z}=\frac1w\times\frac1{1-\frac zw}=\frac1w+\frac z{w^2}+\frac{z^2}{w^3}+\cdots$$and since $$\frac1{(w-z)^2}=\left(\frac1{w-z}\right)',$$you have$$\frac1{(w-z)^2}=\frac1{w^2}+\frac{2z}{w^3}+\frac{3z^2}{w^4}+\cdots$$

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  • $\begingroup$ Thanks for your hint! Why are we allowed to use the geometric sum here? Do we know anything about the ratio $\frac{z}{w}$ (especially that it is $<1$?) I added some steps, following from your hint. Did I understand this right? I'm not sure, wether the result is consistent with Cauchy's residue formula(?) $\endgroup$ – pcalc Apr 9 '18 at 19:28
  • $\begingroup$ @pcalc I can't answer your questions. Your question is rather vague and I tried to help as best as I can. $\endgroup$ – José Carlos Santos Apr 9 '18 at 20:10

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