# Calculating $\oint_{\gamma}\bar z^ndz$ for a known path

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.

• @user: the previous one has no complex conjugation? May 16, 2019 at 13:50
• @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) May 16, 2019 at 13:55

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 $$0$$).
• Let $\eta(t)=1-t+ti$. Then$$\int_\eta\overline z^n\,\mathrm dz=\int_0^1\overline{1-t+ti}^n(-1+i)\,\mathrm dt=(-1+i)\int_0^1(1-t-ti)^n\,\mathrm dt.$$ May 16, 2019 at 14:45