Curve integrals, Polynomial conjugated, inserting e^(it) inverse

I attempted to solve these questions from old examination papers, the first I could do, whether they are correct or not I don't know, and the following I am sure I got up with the wrong foot to begin with:

1. Compute a) $\int_{\partial B_{r}(0)} \overline{z} dz$; b) $\int_{\partial B_{r}(0)}Re(z)dz$
2. Prove that : if p is y polynomial, then $\int_{\partial B_{r}(0)} \overline{p(z)}dz = 2\pi r^{2}\overline{p'(0)}$
3. Prove that : $\int_{\partial B_{1}(0)}f(z)dz = \int_{\partial B_{1}(0) }f(\frac{1}{z})\frac{dz}{z^{2}}$

1. a) I used this parametrization : $re^{it}, t\in [0,2\pi]$ and get the following $\int_{0}^{2\pi}re^{-it} ire^{it}dt = ir^{2}\int_{0}^{2\pi}1dt = 2\pi ir^{2}$ b) $\int_{0}^{2\pi} Re( re^{it}) dz = \int_{0}^{2\pi} rcos(t) ire^{it}dt = ir^{2}\int_{0}^{2\pi}cos(t)e^{it}dt = 2\pi ir^{2}$
2. I put p as polynomial of degree n : $z^{n}+z^{n-1}...+z_{0}$, then if we insert the parametrization we get: $\int_{0}^{2\pi} \overline{(r^{n}e^{nit}+r^{n-1}e^{(n-1)it}+...+z_{0}})(ire^{it})dt$ $= ir\int_{0}^{2\pi}(r^{n}e^{(-n+1)it}+r^{n-1}e^{(-n-1+1)it} + r^{n-2}e^{(-n-1)it}+... \overline{z_{0}})dt$

$=ir\int_{0}^{2\pi}r^{n}e^{(-n+1)it}dt + ir\int_{0}^{2\pi} r^{n-1}e^{-nit}dt ... + ir\int_{0}^{2 \pi}\overline{z_{0}}dt$

3 ) $\int_{\partial B_{1}(0)}f(z)dz = \int_{0}^{2\pi}f(e^{it})ie^{it}dt$ then insert here something and $=\int_{0}^{2\pi} f(e^{-it})e^{-2it}ie^{it}dt = \int_{\partial B_{1}(0)}f(z^{-1})z^{-2}dz$

How to fill in the missing parts? Are my beginnings OK? Would be very glad if somebody would help me out . Thanks for every effort.

(Sorry if this is a little bit much at once. )

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There is no reason why you shouldn't be able to use your method in principle. However, I think you are making life too difficult for yourself by going back to real integration instead of using the magic of complex numbers!

Recall that $z\bar{z}=|z|^2$. So you can write $\bar{z}=|z|^2/z$, and use what you should already know about $$\int_{\partial B_r(0)} z^m dz$$ (where $m\in\mathbb{Z}$) to evaluate $$\int_{\partial B_r(0)} \bar{z}^{-m} dz.$$ This will allow you to solve 1a) and 2. (Note that, in your answer, you have an extra factor of $2$ that shouldn't be there.)

For 1b), just remember that $z+\bar{z}=2\operatorname{Re}(z)$.

Finally, for 3., consider the transformation $w=1/z$. Then $dw=-dz/z^2$, and this transformation takes the unit circle $\partial B_1(0)$ to $\overline{\partial B_1(0)}$, by which I mean the unit circle traversed in negative orientation. So

$$\int_{\partial B_1(0)} f(w)dw = \int_{\overline{\partial B_1(0)}} -f(1/z) \frac{dz}{z^2} = \int_{\partial B_1(0)} f(1/z) \frac{dz}{z^2}.$$

Of course to formalize this, it might actually be easiest to use your parametrization. To do so, just use the transformation $t=2\pi -u$: $$\int_{0}^{2\pi} f(e^{it})i e^{it}dt = \int_{2\pi}^0 f(e^{-iu})ie^{-iu}(-du) = \int_0^{2\pi} f(e^{-iu})ie^{-iu}du.$$

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