Prove that the multiplicative groups $\mathbb{R} - \{0\}$ and $\mathbb{C} - \{0\}$ are not isomorphic.

Is my proof correct? I have made use of the fact isomorphism preserves order of elements, which I proved couple of exercises back. I am also interested in other ways of proving it. Is there a more explicit way or is this explicit enough?

Problem Prove that the multiplicative groups $\mathbb{R} - \{0\}$ and $\mathbb{C} - \{0\}$ are not isomorphic.

Solution Recall that isomorphism preserves order of elements and hence if there exists an isomorphism from $\phi: \mathbb{C}-\{0\} \mapsto \mathbb{R}-\{0\}$, then $x \in \mathbb{C} - \{0\}$, and $\phi(x) \in \mathbb{R} - \{0\}$, then $\vert x \vert = \vert \phi(x) \vert$. Now note that the element $i \in \mathbb{C} - \{0\}$ has order $4$. However, no element in $\mathbb{R}-\{0\}$ has order $4$. Hence, no isomorphism can exist. Hence, the multiplicative groups $\mathbb{R} - \{0\}$ and $\mathbb{C} - \{0\}$ are not isomorphic.

Thanks

Another way would be. Let's take $i$ and see what happen:
a) $f(i^2)=f(-1)=-1$, using basic properties of a morphism.
b) on the other side, we have $f(i)^2=x^2$ that can't produce -1.