- #1
futurebird
- 272
- 0
I'm trying to show that
[tex]\int_{c}z^{n}dz= \left\{\frac{0, n\neq-1}{2\pi i, n=-1}\right[/tex]
I did a change of variables with [tex]z=e^{i\theta}[/tex] and [tex]dz=ire^{i\theta}d\theta[/tex]:
[tex]=i\int^{2\pi}_{0}r^{n+1}e^{i(n+1)\theta}d\theta[/tex]
[tex]=ir^{n+1}\int^{2\pi}_{0}e^{i(n+1)\theta}d\theta[/tex] Moving the constant out.
[tex]=-(n+1)r^{n+1}\int^{2\pi}_{0}\frac{e^{i(n+1)\theta}d\theta}{i(n+1)}[/tex] Getting ready to integrate.
[tex]=-r^{n+1}(n+1)\left[e^{i(n+1)\theta}\right]^{2\pi}_{0}[/tex]
[tex]=-r^{n+1}(n+1)[/tex]?
This is nothing like the answer... where am I going wrong?
[tex]\int_{c}z^{n}dz= \left\{\frac{0, n\neq-1}{2\pi i, n=-1}\right[/tex]
I did a change of variables with [tex]z=e^{i\theta}[/tex] and [tex]dz=ire^{i\theta}d\theta[/tex]:
[tex]=i\int^{2\pi}_{0}r^{n+1}e^{i(n+1)\theta}d\theta[/tex]
[tex]=ir^{n+1}\int^{2\pi}_{0}e^{i(n+1)\theta}d\theta[/tex] Moving the constant out.
[tex]=-(n+1)r^{n+1}\int^{2\pi}_{0}\frac{e^{i(n+1)\theta}d\theta}{i(n+1)}[/tex] Getting ready to integrate.
[tex]=-r^{n+1}(n+1)\left[e^{i(n+1)\theta}\right]^{2\pi}_{0}[/tex]
[tex]=-r^{n+1}(n+1)[/tex]?
This is nothing like the answer... where am I going wrong?