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Let $C_R$ be the positively oriented circle with centre $3i$ and radius $R > 0$. Use the Cauchy Residue Theorem to evaluate the integral $$\int_{C_R}\frac{z^3}{(z-1)(z-4)^2}$$ Your answer should state any values of R for which the integral cannot be evaluated.

Now I can find the residues of the function easily enough. They are $\frac{1}{9}$ at $z=1$ and $\frac{80}{9}$ at $z=4$. However I have no idea at what radius the points $z=1$ and $z=4$ are included in the region being integrated. Is there a way to calculate the radius at which these points need to be taken into account?

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Sketch 3i , 1, 4 on a paper and use pythagorean theorem.

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  • $\begingroup$ No sorry my question was, as the radius is undetermined in the question, we will have different answers depending on how large the radius is. My question is what is the limit the radius has to be to include say the first point but not the second $\endgroup$ – George1811 May 16 '15 at 22:32
  • $\begingroup$ Oh, sketch 3i , 1, and 4 on a paper and use pythagorean theorem $\endgroup$ – grdgfgr May 16 '15 at 22:33
  • $\begingroup$ Ah, clearly of course $\endgroup$ – George1811 May 16 '15 at 22:35

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