How to calculate
$$\int {sin z \over ( z + 1 ) sin {1\over z}}$$ $ on$ $$| z | = {1\over 2}$$
My attempt was to Binding $\frac{1}{\pi}$ and $0$ by two distinct circle each contains only one trouble point to get
$$\int {sin z \over ( z + 1 ) sin {1\over z}}$$ $ on$ $$|z| = 0.1$$ ..(1)
then multiply by $\frac{{1\over z}}{{1\over z}}$
hence; $$\lim_{z\to 0}{sin z \over ( z + 1 ) sin {1\over z}}$$ is finite
so; ${1\over \pi}$ is removable singularity , therefor
$$\int {sin z \over ( z + 1 ) sin {1\over z}}$$ $ on$ $$|z | = 0.1$$ is equal to $0$
$$\int {sin z \over ( z + 1 ) sin {1\over z}}$$
$on$ $$|z - {1\over \pi} | = 0.1$$ ..(2)