\( \int _ { | z - i | = 1 } \sin z d z \)

To evaluate the integral \( \int _ { | z - i | = 1 } \sin z \, dz \), we can use the fact that \(\sin z\) is an entire function. This means it is analytic everywhere in the complex plane, and hence on any closed contour, the integral of an analytic function is zero according to Cauchy's integral theorem. Since the circle \( |z - i| = 1 \) is a closed contour in the complex plane, we conclude that: \[ \int _ { | z - i | = 1 } \sin z \, dz = 0. \] It's as simple as that! The integral around this circular path evaluates to zero due to the properties of analytic functions.