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[SOLVED] circle radius 2 oriented counterclockwise

dwsmith

Well-known member
Feb 1, 2012
1,673
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

Well-known member
MHB Math Helper
Jan 29, 2012
661
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:

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