Hi I'm having trouble with the following integral
$$ \int_C\left( \frac{sinz}{z+3-i}+\frac{e^z + z^2 - 1}{(z+1)^2} \right)dz $$
Where $ C: |z| = 2$
This is what I have so far.
$$ z + 3 - i = 0$$ $$ z = -3 + i $$ As $ z = - 3 + i $ doesn't lie in the interior of the circle, $$\int_C \frac{sinz}{z+3-i}dz = 0$$
For $$ (z+1)^2 = 0 $$ $$ z = -1 $$
This value of $z$ lies in the interior of the circle. Not sure what to do from here. Thanks.
By CIF:
$$\oint_C\frac{e^z+z^2-1}{(z+1)^2}dz=\left.2\pi i\frac d{dz}\left(e^z+z^2-1\right)\right|_{z=-1}=\left.2\pi i(e^z+2z)\right|_{z=-1}=$$
$$=2\pi i(e^{-1}-2)$$