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

$$left|int_C frac{sinleft(frac{1}{z}right)}{z^2}:dzright| = int_{frac{2}{pi}}^1 frac{sinleft(frac{1}{x}right)}{x^2}:dx = cosleft(frac{1}{x}right)Biggr|_{frac{2}{pi}}^1 = cos 1$$

$mathbf{EDIT}$: Sine $z_1$ was really $i$, the answer becomes

$$|cos(-i)| = cosh 1$$