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.