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}

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}
