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:
- Compute a) $\int_{\partial B_{r}(0)} \overline{z} dz$; b) $\int_{\partial B_{r}(0)}Re(z)dz$
- Prove that : if p is y polynomial, then \int_{\partial B_{r}(0)} \overline{p(z)}dz = 2\pi r^{2}\overline{p'(0)}
- Prove that : $\int_{\partial B_{1}(0)}f(z)dz = \int_{\partial B_{1}(0) }f(\frac{1}{z})\frac{dz}{z^{2}}$
- 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) $\int_{0}^{2\pi} Re( re^{it}) dz = \int_{0}^{2\pi} rcos(t) ire^{it}dt = ir^{2}\int_{0}^{2\pi}cos(t)e^{it}dt = 2\pi ir^{2}$
- 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. )