I need to find $$\left | \int_{C} \frac{\sin{(1/z)}} {z^{2}} dz \right| $$ where $C$ is the straight line from $z_{1} = i$ till $z_{2} = 2/ \pi $.
Now I would very much like some guidance with this. I see we have a singularity at $z_{0} =0 $, but since this is not on the contour nor enclosed by the contour, I would argue this is not a problem. Also, since $C$ is a straight line I reckon we should use parametrization (not sure though). The line segment can be paramtrized by: $$ z(t) = i - \frac{\pi}{2}it,$$ with endpoints $t=0$ and $t=\frac{2}{\pi}$
The if we let $f(z) = \frac{\sin{(1/z)}} {z^{2}} $ and we use the fact that $$\int_{C} \frac{\sin{(1/z)}} {z^{2}} dz = \int_{t_{start}}^{t_{end}} f(z(t)) z'(t) dt $$ we find
$$ \int_{C} \frac{\sin{(1/z)}} {z^{2}} dz = \int_{C} \frac{\sin{(1/(i - \frac{\pi}{2}it))}} {(i - \frac{\pi}{2}it)^{2}} dz, $$
but this to me makes the integral only more troublesome. So I reckon I took a wrong turn somewhere.
The function has an explicit antiderivative
$$\left|\int_C \frac{\sin\left(\frac{1}{z}\right)}{z^2}\:dz\right| = \int_{\frac{2}{\pi}}^1 \frac{\sin\left(\frac{1}{x}\right)}{x^2}\:dx = \cos\left(\frac{1}{x}\right)\Biggr|_{\frac{2}{\pi}}^1 = \cos 1$$
$\mathbf{EDIT}$: Sine $z_1$ was really $i$, the answer becomes
$$|\cos(-i)| = \cosh 1$$