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I have recently learned how to compute complex integrals using the Cauchy integral formula but I have come across one with an nth power involved where I am unsure how to proceed. The integral is:

$$\int_{C_1(i)}\frac{z^3}{(z-i)^n}dz$$

For reference, my definition provided for the Cauchy integral formula is:

$$\bbox[yellow] {f(z_0) = \frac{1}{2πi}\oint_{C_r(\hat z)}{\frac{f(z)}{z-z_0}}dz } $$

EDIT: Have asked my lecturer and apprently it is possible to use Cauchy's integral formula for derivatives but I have had no luck, any help?

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  • $\begingroup$ the integral is equal to $$ 2\pi i D^{n-1}z^{3}$$ $\endgroup$ – Jose Garcia Apr 19 '18 at 15:31
  • $\begingroup$ @JoseGarcia Can you show some working? Unfortunately a simple answer isn't very helpful. $\endgroup$ – Zombiegit123 Apr 19 '18 at 15:33
  • $\begingroup$ Do you know Cauchy's formula for derivatives? $\endgroup$ – Nitin Uniyal Apr 19 '18 at 15:33
  • $\begingroup$ @Mathlover Yes but I am not permitted to use it for this question, only the definition. $\endgroup$ – Zombiegit123 Apr 19 '18 at 15:35
  • $\begingroup$ Are you permitted to use Laurent's Theorem? Specifically the part about $c_n = \dfrac{1}{2\pi i}\int_C \dfrac{f(z)}{(z-z_0)^{n+1}}dz$ ? $\endgroup$ – Andy Walls Apr 19 '18 at 16:00
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Notice that $f(z)=z^3$ is entire. So it is differentiable on $D_{1+\epsilon}(i)$.

Cauchy's integral formula for derivatives tells us that for all $\zeta \in D_{1+\epsilon}(i)$ $$f^{(n-1)}(\zeta)=\frac{(n-1)!}{2\pi i}\int_{C_1(i)}\frac{f(z)dz}{(z-i)^n}$$

Hence $$\int_{C_1(i)}\frac{z^3dz}{(z-i)^n}=\frac{2\pi i f^{(n-1)}(i)}{(n-1)!}$$

Hopefully that is enough for you to determine the integral for various values of $n$

Edit: Set $n'=n-1$ then Cauchy's formula for derivatives states: $$f^{(n')}(\zeta)=\frac{n'!}{2\pi i}\int_{C_1(i)}\frac{f(z)dz}{(z-i)^{n'+1}}$$ Work out this integral and then replace $n'$ with $n-1$ in the final answer

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  • $\begingroup$ Yeah I came to this result but I wasn't sure if you could change the formula from $f^n(i)$ to $f^{n-1}(i)$. I will try this now. $\endgroup$ – Zombiegit123 Apr 24 '18 at 15:19
  • $\begingroup$ You could just set $n'=n-1$ then you have exact Cauchy's formula in terms of $n'$ and after applying it, just substitute in for your $n$ $\endgroup$ – pureundergrad Apr 24 '18 at 15:21
  • $\begingroup$ Do you think you can try this for me? I understand what you mean but can't apply it. Sorry for being difficult. $\endgroup$ – Zombiegit123 Apr 24 '18 at 15:33
  • $\begingroup$ Sure, please see edit $\endgroup$ – pureundergrad Apr 24 '18 at 15:36
  • $\begingroup$ @pureundergsgrad Should it not be $\frac{n!}{2πi)$ at before the integral, instead? $\endgroup$ – Zombiegit123 Apr 24 '18 at 15:41

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