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Good time of day everyone,

So I have this question: Let $C_r$ be the circle of radius $|z-a|=r$. Let P(z) be a polynomial. I need to show that the following integral along $C_r$ is $\int P(z) d\overline{z}=-2\pi r^2 P'(a)$. I do not know how to start this, I do not understand what integration along conjugate of $dz$ is. If this was regular $dz$ I would get zero since Polynomials have no poles... So I'm confused on how to handle this beast. Thank you in advance.

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    Some other answers to this question: (1) https://math.stackexchange.com/q/38481/121988 (2) https://math.stackexchange.com/q/509622/121988 – 2018-02-14

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Note that $(z-a)\left ( \bar{z}-\bar{a} \right )=r^2$. Differentiate this to get

$$d \bar{z} = -\frac{\bar{z}-\bar{a}}{z-a} dz = -\frac{r^2}{(z-a)^2} dz $$

Thus,

$$\oint_{|z-a|=r} d\bar{z} \, P(z) = -r^2 \oint_{|z-a|=r} dz \frac{P(z)}{(z-a)^2} $$

Now use Cauchy's integral theorem.