For $\theta \in \mathbb{R}$, define cis($\theta$) = $\cos (\theta ) +i\sin (\theta )$
Show cis: $\mathbb{R}\rightarrow \left \{ z\in \mathbb{C}:\left | z\right | = 1 \right \}$ is a homomorphism
Show that ker(cis) = $2\pi\mathbb{Z} = \left \{ 2\pi n\in \mathbb{R}:n\in\mathbb{Z} \right \}$
I already showed its a homomorphism by:
$$cis(a+b) = cis(a) *cis(b)$$ $$\cos (a+b) +i\sin (a+b ) = (\cos (a) +i\sin (a)) * (\cos (b) +i\sin (b))$$ $$\cos (a+b) +i\sin (a+b ) = [\cos (a)\cos(b) - \sin(a)\sin(b)] +i[\sin (a)\cos(b)+\sin(b)\cos(a)]$$ $$\cos (a+b) +i\sin (a+b ) = \cos (a+b) +i\sin (a+b )$$ I need help showing the kernal, I'm a bit confused on on how to show cis$(2\pi\mathbb{Z}) = e$, $e$ being the identity