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I'm trying to evaluate $\int_{z_1}^{z_2}\bar z^ndz$ along the straight line $[z_1,z_2]$, where $n \in \mathbb{N}\cup \{0\}$ (I found this problem here).

As has been suggested before there, I've tried to use the linear curve $$z=tz_1+(1-t)z_2, 0\leq t\leq 1$$

which leads to $$\int_{z_1}^{z_2}\bar z^ndz=(z_1-z_2)\int_{0}^{1}(t\bar z_1+(1-t)\bar z_2)^ndz$$

but couldn't get any further.

Any help will be greatly appreciated.

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  • $\begingroup$ on the RHS you are integrating from t=0 to t=1 so you should not have $dz$ but $dt$ because $dz=(z_1-z_2)dt.$ $\endgroup$ – DanielWainfleet Nov 19 '15 at 10:00
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Use the binomial theorem:

$$(\bar{z_1} t + \bar{z_2} (1-t))^n = \sum_{k=0}^n \binom{n}{k} \bar{z_1}^k \bar{z_2}^{n-k} t^k (1-t)^{n-k} $$

Thus, the integral is

$$(z_2-z_1)\sum_{k=0}^n \binom{n}{k} \bar{z_1}^k \bar{z_2}^{n-k} \int_0^1 dt \, t^k (1-t)^{n-k} = (z_2-z_1)\frac1{n+1} \bar{z_2}^n \sum_{k=0}^n \left (\frac{\bar{z_1}}{\bar{z_2}} \right )^k = \frac1{n+1} (\bar{z_2}^{n+1}-\bar{z_1}^{n+1}) \frac{z_2-z_1}{\bar{z_2}-\bar{z_1}}$$

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  • $\begingroup$ How did you get $(z_2-z_1)\frac1{n+1} \bar{z_2}^n \sum_{k=0}^n \left (\frac{\bar{z_1}}{\bar{z_2}} \right )^k $? $\endgroup$ – Brassican Nov 19 '15 at 8:57
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    $\begingroup$ @Brassican: The integral is a beta function: $$\int_0^1 dt \, t^k (1-t)^{n-k} = \frac{k! (n-k)!}{(n+1)!}$$ $\endgroup$ – Ron Gordon Nov 19 '15 at 11:26
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When a function $f:C\to C$ has a continuous complex derivative $f'$, and you restrict its domain to $R$, it is easily shown that for $t\in R$ we have $$Re (f'(t))=\frac {d Re (f(t))}{dt} \land Im f'(t)=\frac {d Im (f(t))}{dt}$$ and that these are continuous in $t$. Therefore $$\int_0^1f'(t)dt=\int_0^1Re (f'(t))dt+i\int_0^1Im (f'(t))dt= \int_0^1\frac {dRe (f(t))}{dt}dt+i\int_0^1\frac {dIm (f(t))}{dt}dt$$ $$=[Re (f(t))]_{t=0}^{t=1}+i[Im (f(t))]_{t=0}^{t=1} = f(1)-f(0).$$ In your question, we can let $f'(z)=(z \bar z_1+(1-z)\bar z_2)^n$ so (obviously) we can let $f(z)=(n+1)^{-1}(\bar z_1-\bar z_2)^{-1}(z\bar z_1+(1-z)\bar z_2)^{n+1}$. And the rest is obvious.

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