Application of First Isomorphism theorem - Help with finding Kernel

Question

Are the groups $\Bbb{R}/\Bbb{Z}$ and $S^1$ $isomorphic$? Here $\Bbb{R}$ is the additive group of real numbers, and $S^1 = \{z\in\Bbb{C}\mid |z| = 1\}$ under complex multiplication.

I am not able to define the kernel of the homomorphism.

My attempt:

Consider $f$ from $\Bbb{R}$ to $S^1$ defined as $f(x) = e^{ix}$. Then $f$ is a homomorphism. So by first isomorphism theorem, $\Bbb{R}/\ker f$ is isomorphic to $f(\Bbb{R})$. But $f(\Bbb{R}) = S^1$. So, $\Bbb{R}/\ker f$ is isomorphic to $S^1$.

Now, $\ker f =\{ x\mid f(x) = 1\} = \{ x | e^{ix} = 1\}$. I am not able to proceed further.

How do I find all $x$ that satisfy $e^{ix} = 1$? I know $e^(ix) = \cos x + i\sin x$. So one value of $x$ must be $0$. How do I find the other values of $x$? And does $\ker f$ equal $\Bbb{Z}$?

I know how the set of complex roots of unity for all $n\ge1$ is defined. Is there a way to use that information here?

Answer 1

The $\ker f=\{x\in\Bbb{R}\mid e^{ix}=1\}$ implies that $\cos x+i\sin x=1$, so $\cos x=1$ and $\sin x=0$ simultaneously . Hence $x=2k\pi$ for all $k\in\Bbb{Z}$.

Answer 2

the exponential function is periodic, with period $2\pi i$, so your map $f$ has kernel the additive group $\{2\pi n \mid n \in \mathbb{Z}\} \cong \mathbb{Z}$ (as an additive group)

Answer 3

By definition

$$x\in\Bbb R\implies e^{ix}=\cos x+i\sin x\implies \ker f=\{2k\pi\;:\;\;k\in\Bbb Z\}$$

I think your map would be better defined as $\;f(x):=e^{2\pi ix}\;$ , so that now the kernel is simply $\;\Bbb Z$