$$\int_{\gamma=(i,1)} \frac{z^3}{(z-i)^n} dz$$ for any $n\in\mathbb{N}$.

Can someone please help me answer this question as I cannot seem to get the right answer!

Please note that the Cauchy integral formula must be used in order to solve it.

Many thanks in advance!

  • $\begingroup$ What is your contour? It doesn't make sense currently. $\endgroup$ – anon271828 Mar 15 '13 at 0:24
  • $\begingroup$ well that is exactly how the question appears in the book. $\endgroup$ – camilla Mar 15 '13 at 0:27
  • $\begingroup$ What book are you using? $\endgroup$ – anon271828 Mar 15 '13 at 0:27
  • 1
    $\begingroup$ Maybe it means a circle of radius 1 centered at $i$? $\endgroup$ – Chris Brooks Mar 15 '13 at 0:29
  • $\begingroup$ yes it means |z| = 1 and centred at i. i dont know what a contour is, thats why im asking for help im massively stuck!! $\endgroup$ – camilla Mar 15 '13 at 0:30

Use the residue theorem. Since $\cfrac{z^3}{(z-i)^n}$ has a pole of order $n$ at $z=i$ and analytic everywhere other than $z=i$ in the domain $|z-i|<1$, by residue theorem, we have $$\int_{\gamma}\frac{z^3}{(z-i)^n}dz=2\pi ig(i),\text{where }g(z)=\frac{1}{(n-1)!}(z^3)^{(n-1)}$$

The residue theorem is obtained from Cauchy Integral formula.

By Cauchy Integral formula, we have $$2\pi if(z)=\int_C\frac{f(\zeta)}{\zeta-z}d\zeta$$ Differentiate both sides with respect to $z$ $n-1$ times, we get $$2\pi if^{(n-1)}(z)=(n-1)!\int_C\frac{f(\zeta)}{(\zeta-z)^n}d\zeta$$

  • $\begingroup$ I can't use the residue theorem is has to be the Cauchy Integral Formula!! Sorry but thank you :) $\endgroup$ – camilla Mar 15 '13 at 0:45
  • $\begingroup$ @camilla Residue theorem and Cauchy Integral Formula are actually the same thing. Or, just attach the "two-line proof" to your homework $\endgroup$ – NECing Mar 15 '13 at 0:48
  • $\begingroup$ is there not a way of putting it into partial fractions to solve? this is what im used to - not understanding very well sorry!! btw thank you for your answers. $\endgroup$ – camilla Mar 15 '13 at 0:49
  • $\begingroup$ Contour integral is different from the real one. You can not directly take the anti-derivative. $\endgroup$ – NECing Mar 15 '13 at 0:51
  • $\begingroup$ okay, im a bit confused about this one, but would the answer be (2pi)i? $\endgroup$ – camilla Mar 15 '13 at 0:58

Hint: Parametrize the contour as $z=i+e^{i\theta}$ for $0\leq\theta\leq 2\pi$. Then, by definition, you have $$\int_0^{2\pi}\frac{(e^{i\theta})^3}{((i+e^{i\theta})-i)^n}\cdot ie^{i\theta}\,d\theta=i\int_0^{2\pi}\frac{(e^{i\theta})^3}{e^{in\theta}}\cdot e^{i\theta}\,d\theta=i\int_0^{2\pi}(e^{i3\theta})e^{i(1-n)\theta}\,d\theta.$$Keep simplifying and you can solve it.

  • $\begingroup$ is that using the cauchy integral forumula though?? Many thanks $\endgroup$ – camilla Mar 15 '13 at 0:37
  • $\begingroup$ should i not put it into partial fractions and use the cauchy formula twice? I have a different looking answer here! $\endgroup$ – camilla Mar 15 '13 at 0:42

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.