Wanted to check if I got the right answer/ idea for this question:
$$\int_{|z|=1} \frac{\sin(z)}{z}\mathrm dz$$
Attempt: The region of the curve is the unit circle so there is a singularity at the origin.
Using Cauchy's formula: $$2\pi i \sin(0) = 0 $$
Is this the correct answer as well as the correct way to do this integral?
Yes, that is correct.
You can also say that, since $\lim_{z\to0}\dfrac{\sin z}z=1$, $0$ is a removable singularity of $\dfrac{\sin z}z$. It follows therefore from Cauchy's integral theorem that your integral is equal to $0$.