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When dealing with a contour along a unit circle, we can set $|z| = 1$ and $z(t) = e^{it}$ so that $\frac{d z}{dt} = ie^{it}$ with $t\in [0,2\pi]$.

Find the integral of:

$$\int_{\Gamma}(z^7+z^4) dz$$

My own attempt.

$$\int_{\Gamma} (z^7+z^5)dz = \int_0^{2\pi}((e^{it})^7+(e^{it})^5)(ie^{it})dt $$ $$i\int_0^{2\pi}e^{8it}+e^{6it}dt$$

Am I doing it right? It doesn't seem like the right answer but I can't find anything that says it's not. How would it be done using the Residue Theorem?

mr_e_man
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Chris Cross AppleSause
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  • Have you studied Cauchy's Theorem? You are also applying the definition of contour integral wrongly. – geetha290krm Mar 30 '23 at 09:21
  • Two comments. First, an integral expression $\int\cdots dz$ has implied brackets, $\int(\cdots),dz$. In your first step it looks like you are thinking of the integral as $\int z^7+(z^5,dz)$, but it should be $\int(z^7+z^5)dz$. Second, you may need to revise power laws: $(e^{it})^5(ie^{it})=e^{5it},ie^{it}=ie^{6it}$. – David Mar 30 '23 at 09:27
  • David - Thanks for pointing it out and i agree completely.

    greeth290krm - I have, and i can see that now. Thanks.

     – Chris Cross AppleSause Mar 30 '23 at 09:41
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    To use residue theorem, you just need to see that there are no residues as polynomials are holomorphic functions . Hence the integral is $0$. – Mr.Gandalf Sauron Mar 30 '23 at 09:48

1 Answers1

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No, it is not correct. It should be\begin{align}\int_\Gamma z^7+z^5\,\mathrm dz&=\int_0^{2\pi}\left(\left(e^{it}\right)^7+\left(e^{it}\right)^5\right)ie^{it}\,\mathrm dt\\&=i\int_0^{2\pi}e^{8it}+e^{6it}\,\mathrm dt\\&=i\int_0^{2\pi}\cos(8t)+i\sin(8t)+\cos(6t)+i\sin(6t)\,\mathrm dt\\&=i\left[\frac{\sin(8t)}8+i\frac{-\cos(8t)}8+\frac{\sin(6t)}6+i\frac{-\cos(6t)}6\right]_{t=0}^{t=2\pi}\\&=0.\end{align}

José Carlos Santos
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