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.


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?

| cite | improve this question | | | | |

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$$\frac1{(w-z)^2}=\frac1{w^2}+\frac{2z}{w^3}+\frac{3z^2}{w^4}+\cdots$$

| cite | improve this answer | | | | |
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.