My complex class now finished Cauchy Integral Formula, Reflection Principle, and Residue Formula, and I am asked to show:
Let $P(z)$be a polynomial and let $C$ be the circle $\mid z-a\mid$=$R$.
Show $\int_{C}$$P(z)d\bar{z}=-2\pi iR^{2}P^{'}(a)$.
Here is my thought:
Since it related to $2\pi i$ and $P^{'}(a)$, I think it must be related to Cauchy Integral Formula, also since we have to integral respect to the conjugate of $z$, and the circle is symmetric respect to the real axis, I think it may be related to the Reflection Principle
So I let $P(z)=a_{n}z^{n}+\cdots+a_{1}z+a_{0}$, since $P(z)$ is a polynomial. Then, we have to deal with different cases about $R$.
If $R\geq a$, then $0$ is inside $C$, we can apply Cauchy Integral Formula.
Let $f(\zeta) = a_{n}\zeta^{n+1}+\cdots+a_{1}\zeta^{2}+a_{0}\zeta$.
Then $\int_{C}$$P(z)d{z}$ = $\int_{C}$$\frac{f(\zeta)}{\zeta-0}$ = $2\pi i f(0)=0$
And I don't know how to proceed further. This result is not right since I did not even consider the conjugate of $z$, since I don't know how to consider it.
Also, I don't know how to deal with the case $R\leq a$, since in that case, Cauchy Integral Formula is not this easy to use again.
Any hints and detailed explanations are really appreciated! Also, this question is due next Wednesday, and we have class on Monday, so if you think I should use other theorems, please feel free to tell me!
$|z-a|^{2}=R^{2}$, $(z-a)(\overline{z}-\overline{a})=R^{2}$, so $d(z-a)(\overline{z}-\overline{a})=0$, or $(\overline{z}-\overline{a})dz+(z-a)d\overline{z}=0$, so $d\overline{z}=-\dfrac{\overline{z-a}}{z-a}dz$, hence \begin{align*} \int_{C}P(z)d\overline{z}&=-\int_{C}P(z)\dfrac{\overline{z-a}}{z-a}dz\\ &=-\int_{0}^{1}P'(a)(Re^{2\pi i\theta})e^{-4\pi i\theta}Re^{2\pi i\theta}i2\pi d\theta\\ &=-2\pi iR^{2}P'(a), \end{align*} where $z=a+Re^{2\pi i\theta}$ is the parametrized curve.