use cauchy gaussat theorem to evaluate the integral of

Question

(z-3)/(z²+2z+2)

Comments

The contour has not been specified and the integral for Cauchy-Goursat hasn't been included. 

Answer 1

z²+2z+2 can be factorised:

z²+2z+2=z²+2z+1+1=(z+1)²+1=(z+1)²-i²=(z+1-i)(z+1+i).

(z-3)/((z+1-i)(z+1+i))=(A+Bi)/(z+1-i)+(C+Di)/(z+1+i) (partial fractions).

Az+A+Ai+Biz+Bi-B+Cz+C-Ci+Diz+Di+D=z-3,

Az+Cz=z, A-B+C+D=-3, B+D=0, A+B-C+D=0

-B+D=-4; B+D=0⇒D=-2, B=2; A-C=0, A+C=1⇒A=C=½.

(z-3)/(z²+2z+2)=(½+2i)/(z+1-i)+(½-2i)/(z+1+i).

CHECK:

(½+2i)(z+1+i)+(½-2i)(z+1-i)=

½z+½+½i+2iz+2i-2+½z+½-½i-2iz-2i-2=z-3✔️

Let J=∫((z-3)/(z²+2z+2))dz=(½+2i)∫dz/(z+1-i)+(½-2i)∫dz/(z+1+i).

Apply ∫(f(z)/(z-z₀)=2πif(z₀) (Cauchy Integral)

We can apply this to the two integrals separately:

In each case f(z)=1=f(z₀), independent of z₀.

so we have J=(½+2i)(2πi)+(½-2i)(2πi)=πi-4π+πi+4π=2πi.

Note that the region defined by the unspecified contour must contain z₀ in each case.