Let $R$ be the rectangle with vertices at $-1$, $1$, $1+2i$, $-1+2i$. Compute $$\int_{\partial R} \frac{(z^2 +i)\sin(z)}{z^2+1}dz$$where the boundary of $R$ is traversed counterclockwise.
Here is what I got. Let $$f(z)=\frac{(z^2 +i)\sin(z)}{z^2+1}dz$$ I then factor the bottom to get $$f(z)=\frac{(z^2 +i)\sin(z)}{(z-i)(z+1)}dz$$
Now i can use Cauchy-Goursat integral formula. Since $z=i$ lies in the rectangle thus the answer will be $2\pi i$ times something which I do not know and when $z=-i$ well the integral goes to zero. Now let us get back to when $z= i$ then the integral goes to $2\pi i$ times constant. now to find that constant i need to use partial fraction thus I have $$\frac{(z^2 +i)\sin(z)}{(z-i)(z+1)} = \frac {A}{z-i}+\frac{B}{z+i}$$
We now get $$(z^2 +i)\sin(z) = A(z+i) +B(z-i)$$
If $z=i$ then $$A=\frac{(-1+i)\sin(i)}{2i}$$which will give me an answer of $$- \pi + \pi i$$ when I plug it back in $$\frac {A}{z-i}$$
I feel like things are not what they supposed to be so Now my question is am I in the right path of this or am I doing it bad.
The rectangle encloses the pole at $z=i$. Hence, the integral is $$2 \pi i \left( \text{Residue of $f(z)$ at }z=i\right)$$ The residue of $f(z)$ at $i$ is given by $\lim_{z \to i} (z-i)f(z)$. We have $f(z) = \dfrac{z^2+i}{z^2+1} \sin(z)$. Hence, $$\lim_{z \to i} (z-i) \cdot \dfrac{z^2+i}{z^2+1} \sin(z) = \lim_{z \to i} \dfrac{z^2+i}{z+i} \sin(z) = \dfrac{i^2+i}{2i} \sin(i) = \dfrac{i+1}2 \sin(i) = \dfrac{i-1}2 \sinh(1)$$