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I'm trying to evaluate $$\int_{|z|=2}\frac{e^{3z}}{(z-1)^3}dz$$ using the residue theorem. I get a pole of order $3$ at 1 with a residue of $\frac{9}{2}e^3$. But since the absolute value of the residue (which in this case is exactly the residue) is bigger than 2, does that mean the integral is equal to zero? I'm just really confused, since this is the first in a series of practice problems and I was expecting to have to use the residue theorem.

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    $\begingroup$ Why does the absolute value of the residue matter? $\endgroup$ May 24 '20 at 22:28
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    $\begingroup$ Thanks for clearing it up, of course only the singularity has to be within the bounds of the integral, not the residue. $\endgroup$
    – MJP
    May 24 '20 at 22:33
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The absolute value of the residue is irrelevant here. Since$$\operatorname{res}\left(1,\frac{e^{3z}}{(z-1)^3}\right)=\frac92e^3,$$then$$\int_{|z|=2}\frac{e^{3z}}{(z-1)^3}\,\mathrm dz=9e^3\pi i.$$

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  • $\begingroup$ Thank you, I wrongly thought that the residue has to be within the bounds. Should I remove the question since it doesn't really raise any issue? $\endgroup$
    – MJP
    May 24 '20 at 22:35
  • $\begingroup$ Why? You had a doubt and now, thanks to my answer, you don't have it anymore. $\endgroup$ May 24 '20 at 22:40
  • $\begingroup$ Of course, your answer helped me a lot, I was just concerned that my question might be of low quality since it is just based on me being unable to read the definition of the residue theorem. $\endgroup$
    – MJP
    May 24 '20 at 22:44

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