- #1
fauboca
- 158
- 0
Without using Cauchy's Integral Formula or Residuals, I am trying to integrate
[tex]\int_{\gamma}\frac{dz}{z^2+1}[/tex]
Around a circle of radius 2 centered at the origin oriented counterclockwise.
[tex]\frac{i}{2}\left[\int_0^{2\pi}\frac{1}{z+i}dz-\int_0^{2\pi}\frac{1}{z-i}dz\right][/tex]
[tex]\gamma(t)=2e^{it}, \quad \gamma'(t)=2ie^{it}[/tex]
The answer is zero. I am supposed to get each integral to be [itex]2\pi i[/itex] which is 0 when subtracted.
I know it is related to the fact that [itex]\int_{\gamma}\frac{1}{z}dz = 2\pi i[/itex].
And using u-sub isn't correct since any closed path would be zero when that isn't true. The integral of 1/z shows that not all closed paths will be zero.
[tex]\int_{\gamma}\frac{dz}{z^2+1}[/tex]
Around a circle of radius 2 centered at the origin oriented counterclockwise.
[tex]\frac{i}{2}\left[\int_0^{2\pi}\frac{1}{z+i}dz-\int_0^{2\pi}\frac{1}{z-i}dz\right][/tex]
[tex]\gamma(t)=2e^{it}, \quad \gamma'(t)=2ie^{it}[/tex]
The answer is zero. I am supposed to get each integral to be [itex]2\pi i[/itex] which is 0 when subtracted.
I know it is related to the fact that [itex]\int_{\gamma}\frac{1}{z}dz = 2\pi i[/itex].
And using u-sub isn't correct since any closed path would be zero when that isn't true. The integral of 1/z shows that not all closed paths will be zero.