Evaluate the fallowing integral where in case C is the circle $$\left| z \right|$$ =3, counterclockwise

$$\int\limits_{C}^{}zdz/(z ^{2}-3z+2)$$

Question

Evaluate the fallowing integral where in case C is the circle $$\left| z \right|$$ =3, counterclockwise

$$\int\limits_{C}^{}zdz/(z ^{2}-3z+2)$$

anonymous · Answer

@Herp_Derp can u help me with this too bro ?

anonymous · Answer

i guess i am learning slowly :) i understood the last one i guess

anonymous · Answer

first one was 1/2 like the second one, but when we add them, why we multiplied with 2i

anonymous · Answer

2pi i sorry

anonymous · Answer

Ok, so factor the denominator: $$z^2-3z+2=(z-2)(z-1)$$So there are singularities at z=2 and z=1. Now the circle has a radius of 3, so it encloses both of these singularities, and you use the Cauchy residue theorem. Res(f,1)=1/(1-2)=-1, Res(f,2)=2/(2-1)=2 The sum of the residues is 1, so the final answer is 2πi. We multiplied by 2πi at the end because that is what the theorem says: http://en.wikipedia.org/wiki/Residue_theorem