Operations used in the isomorphism theorem

Question

Prove that $\mathbb{R}/ \mathbb{Z}\cong G=\{a+ib \in \mathbb{C}\;:\; a^2+b^2=1\}.$

In my answer I've tried to show that there is an homomorphis between $\mathbb{R}$ and G. I also tried to show that $Ker(f)=\mathbb{Z}$ and to show that f is surjective. My goal is to use the isomorphism theorem. But I coudn't achieve success.

If I define $f: \mathbb{R} \rightarrow G$ given by $\alpha \longmapsto f(\alpha)= \cos(2\pi \alpha)+ i \sin(2\pi \alpha)$ as the same given as a hint in the Herstein book. I made big mistakes early in the begining. I mean, I'm struggling to show that $\mathbb{Z} = ker(f)$ Let me show you how I wrote:

Let $Kef(f) \subset \mathbb{Z}=\{\alpha\in \mathbb{R}: \; f(\alpha)=1=1+i0\}$. I showed that if $\alpha \in \ker(f) \implies f(\alpha)=1+i0 \implies 1^2+ 0^2=1\in \mathbb{Z}.$

But, how can I proof that $\mathbb{Z}\supset kef(f)$?

Answer 1

$a\in Ker(f)$ iff $f(a)=1$ iff $a$ is an integer and hence $\ker(f)=\mathbb{Z}$.

[Answer for your doubt; Observe that if $f(a)=1$ then $a$ should be an integer. For non integral values of $a,$ $\cos 2\pi a$ will not give you $1.$ ]

Answer 2

If $\cos(2\pi\alpha)+i\sin(2\pi\alpha)=1$, then $\cos(2\pi\alpha)=1\land\sin(2\pi\alpha)=0\implies\alpha\in\Bbb Z$.