Please can somebody give me hint for this?
For $n\ne -1$ $\frac{1}{2\pi i}\int_{C} z^ndz=0$
Where C is a simple closed curve with the usual positive orientation and its inside.
Please can somebody give me hint for this?
For $n\ne -1$ $\frac{1}{2\pi i}\int_{C} z^ndz=0$
Where C is a simple closed curve with the usual positive orientation and its inside.
let my $C(t)= re^{it}$, $0\le t\le 2\pi$ and n is an integer, It follows that $I=r^{n+1} \int_{0}^{2\pi}ie^{(n+1)it}dt= r^{(n+1)}\frac{e^{it(n+1)}}{n+1}|_{0}^{2\pi} \text {when }n\ne -1$ and $2\pi i$ when $n=-1$, that is $\int f(z)dz= 0$ when $n\ne -1$
For $n\neq -1$, $f(z)=z^n$ has a primitive $F(z)=\frac{z^{n+1}}{n+1}$. Parameterize $\gamma$ with $t$ from 0 to 1.
$\displaystyle \int_\gamma f(z) \ dz = \displaystyle \int_\gamma F'(z)\ dz = \displaystyle \int_0^1 F'(\gamma(t)) \gamma'(t)\ dt = \displaystyle \int_0^1 (F(\gamma(t)))'\ dt =F(\gamma(1))-F(\gamma(0))=0$ because the path is closed.
This works even if the curve is not simple.