Show that $f:\mathbb{R}^+ \longrightarrow \mathbb{C}^\times$ defined by $f(x)=e^{ix}$ is a homomorphism Can someone please verify my proof?

Show that $f:\mathbb{R}^+ \longrightarrow \mathbb{C}^\times$ defined by $f(x)=e^{ix}$ is a homomorphism, and determine its kernel and image.

Let $x$ and $y$ be arbitrary elements of $\mathbb{R}^+$. Then,
\begin{eqnarray}
f(x+y) &=& e^{i(x+y)} \\
&=& e^{ix}e^{iy} \\
&=& f(x)\times f(y)
\end{eqnarray}
Also,
\begin{eqnarray}
\operatorname{Im}(f) &=& \{e^{ix}:x \in \mathbb{R}^+\} \\
&=& \{x \in \mathbb{C}: |x|=1\}
\end{eqnarray}
And,
\begin{eqnarray}
\operatorname{ker}(f)&=&\{x \in \mathbb{R}^+:e^{ix}=1\} \\
&=& \{2 \pi n: n \in \mathbb{Z}\}
\end{eqnarray}
 A: Perhaps you mean $(\Bbb R,+)$ and $(\Bbb C^\times,\cdot)$; but as written now there is a problem which bothers me, namely $\Bbb R^+$ often denotes the positive elements of $\Bbb R$, and therefore the kernel of $f$ cannot possibly include negative elements such as $-2\pi$.
Other than that, I don't know what theorems you've seen before, but you haven't really proved what the image and kernel are. You just wrote sets. Why is the kernel exactly those numbers which have the form $2n\pi$ where $n\in\Bbb Z$? and why is the image of $f$ exactly $x\in\Bbb C$ such that $|x|=1$?
(It is correct, though, and it might be that you've seen theorems and exercises before this one, which mitigate the lack of details. I couldn't possibly know.)
A: This question probably originates from Ex 2.4.6 of the book Algebra by Michael Artin.  I believe the notation there

$f:\mathbb{R}^+ \longrightarrow \mathbb{C}^\times$ 

means

$f:(\mathbb{R},+) \longrightarrow (\mathbb{C},\times)$ 

In particular, it doesn't mean the domain is limited to only positive real numbers.
Therefore:
\begin{eqnarray}
\operatorname{Im}(f) &=& \{e^{ix}:x \in \mathbb{R}\} \\
&=& \{x \in \mathbb{C}\}
\end{eqnarray}
and
\begin{eqnarray}
\operatorname{ker}(f)&=&\{x \in \mathbb{R}:e^{ix}=1\} \\
&=& \{2 \pi n: n \in \mathbb{Z}\}
\end{eqnarray}
