MHB Circle radius 2 oriented counterclockwise

[Dustinsfl]

gamma is a circle of radius 2, centered at the origin, and oriented counterclockwise

$\displaystyle\int_{\gamma}\frac{dz}{z^2+1} =\int_{\gamma}\frac{dz}{(z+i)(z-i)}=\frac{1}{2}\int_{\gamma}\frac{\frac{1}{z-i}}{z-(-i)}dz+\int_{\gamma}\frac{\frac{1}{z+i}}{z-i}dz = 4\pi i\left(\frac{1}{-2i}+\frac{1}{2i}\right) = 0$Is this correct?

 

[Fernando Revilla]

gamma is a circle of radius 2, centered at the origin, and oriented counterclockwise

$\displaystyle\int_{\gamma}\frac{dz}{z^2+1} =\int_{\gamma}\frac{dz}{(z+i)(z-i)}=\frac{1}{2}\int_{\gamma}\frac{\frac{1}{z-i}}{z-(-i)}dz+\int_{\gamma}\frac{\frac{1}{z+i}}{z-i}dz = 4\pi i\left(\frac{1}{-2i}+\frac{1}{2i}\right) = 0$ Is this correct?

Right. Only minor mistakes:

$\displaystyle\int_{\gamma}\frac{dz}{z^2+1} =\int_{\gamma}\frac{dz}{(z+i)(z-i)}=\frac{1}{2}\;\left(\int_{\gamma}\frac{\frac{1}{z-i}}{z-(-i)}dz+\int_{\gamma}\frac{\frac{1}{z+i}}{z-i}dz\right) = \frac{1}{2}\cdot 2\pi i\;\left(\frac{1}{-2i}+\frac{1}{2i}\right) = 0$

 

[Greg Bernhardt]

Yes, this is correct. The integral evaluates to 0 because the function has two poles at z = i and z = -i, both of which are inside the circle of radius 2. Therefore, by Cauchy's Integral Theorem, the integral around the entire circle is equal to 0.