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As the title says, I have to compute $$\int_{|z|=r} x^2dz$$ for the circle traversed anti-clockwise in 2 different ways.

If I use a parametrisation I quickly get to 0 - which might be wrong though -, but I'm not really sure how to go about the other way. I got to $x=\frac{1}{2}(z-\frac{r^2}{z})$, but does this help? The only thing I could do is use $re^{it}$ but that would be the very same way as the first one. Any help please?

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  • $\begingroup$ Does $x=\mathrm{Re}(z)$? $\endgroup$
    – user170231
    Nov 10, 2015 at 18:32
  • $\begingroup$ Yes, x should be the real part of z. $\endgroup$
    – Drn004
    Nov 10, 2015 at 18:39
  • $\begingroup$ How about symmetry? $dz$ reverses sign at the antipode. $\endgroup$
    – GEdgar
    Nov 10, 2015 at 19:09
  • $\begingroup$ I'm not following.. could you please explain a little bit? $\endgroup$
    – Drn004
    Nov 10, 2015 at 21:53

1 Answer 1

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Setting $z=re^{it}$, you have $$\oint_{|z|=r}\mathrm{Re}(z)^2\,\mathrm{d}z=ir^3\int_0^{2\pi}\cos^2t\,e^{it}\,\mathrm{d}t=\frac{ir^3}{2}\int_0^{2\pi}\bigg(1+\cos2t\bigg)e^{it}\,\mathrm{d}t$$ which agrees with your other approach, giving an answer of $0$.

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  • $\begingroup$ Thanks, but basically this is my approach, I need another one $\endgroup$
    – Drn004
    Nov 10, 2015 at 19:05
  • $\begingroup$ I'm not sure I follow what you're trying to do, but it's been a while since I dabbled in complex analysis. You should have $x=\dfrac{z+\bar{z}}{2}$, so that $x^2=\dfrac{z^2+2z\bar{z}+\bar{z}^2}{4}=\dfrac{z^2+\bar{z}^2}{4}+\dfrac{r^2}{2}$, but I'm not seeing how that would help. $\endgroup$
    – user170231
    Nov 10, 2015 at 19:29
  • $\begingroup$ The hint I have for this problem is to rewrite x in terms of z and r, so that's the only way I could do it. I'm not seeing how this could help either, that's why I'm stuck... $\endgroup$
    – Drn004
    Nov 10, 2015 at 19:33

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