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I wish to calculate the following integral:

$$\oint_{\gamma}\bar z^ndz$$

$\gamma$ is a triangle with vertices at $0,1,i$, in the positive direction and $n\in \mathbb Z$.

Since $\bar z^n$ is not analytic, I can't say that the integral is 0, so it has to be calculated manually. I tried to use that $\bar z=x-iy$, but I still could not solve this.

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  • $\begingroup$ @user: the previous one has no complex conjugation? $\endgroup$
    – user10354138
    May 16, 2019 at 13:50
  • $\begingroup$ @user: it could be, depending on how you interpret the question. $\oint_\gamma\bar{z}^n\,\mathrm{d}z\neq 0$, compared with Cauchy implies $\oint_\gamma z^n\,\mathrm{d}z=0$, for $n\geq 0$. So it is really essential to calculate the integral (as opposed to just noting the divergence at $0$ for $n<0$ in the previous case) $\endgroup$
    – user10354138
    May 16, 2019 at 13:55

1 Answer 1

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Let $\eta(t)=t$, with $t\in[0,1]$. Then\begin{align}\int_\eta\overline z^n\,\mathrm dz&=\int_0^1\overline{\eta(t)}^n\eta't\,\mathrm dt\\&=\int_0^1t^n\,\mathrm dt\\&=\frac1{n+1}.\end{align}Now, do the same thing with the other two sides of the triangle (the side that gois from $1$ to $i$ and the side that goes from $i$ to $0$).

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  • $\begingroup$ Can you please show me how to calculate the path from 1 to i? $\endgroup$
    – segevp
    May 16, 2019 at 14:42
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    $\begingroup$ Let $\eta(t)=1-t+ti$. Then$$\int_\eta\overline z^n\,\mathrm dz=\int_0^1\overline{1-t+ti}^n(-1+i)\,\mathrm dt=(-1+i)\int_0^1(1-t-ti)^n\,\mathrm dt.$$ $\endgroup$
    – José Carlos Santos
    May 16, 2019 at 14:45

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