2.4.6

$$f$$ is a homomorphism because $$f(x + y) = e^{i(x + y)} = e^{ix}e^{iy} = f(x)f(y).$$ The kernel is $$2\pi{}n, \forall n \in \mathbb{Z}$$ as this is equivalent to $$e^{ix} = 1$$. The image of $$f$$ is the unit circle in the complex plane.