I'm hopeful that I proved the homomorphism stating:
$$\varphi((x+iy)(a+ib))=(x^2+y^2)(a^2+b^2)=\varphi(x+iy)\varphi(a+bi)\label1\tag1$$
Now I want to show what the kernel and image of $\varphi$ is (I haven't taken complex analysis). I've read that the $\ker\varphi$ is a subgroup of $\mathbb{C}^*$ in this case and that the $\ker\varphi$ is the set of {$x \in G\mid\varphi(x)=e$}. Then that might suggest that $\ker\varphi= \lbrace x, y \in \mathbb{C}^*\mid x^2+y^2=e=1 \rbrace = \left\lbrace \frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}} \right\rbrace$ because of $x^2$ we will always have 1–the identity. Since I think $\mathbb{R}^* \subset \mathbb{C}^*$, then the image of $\varphi$ is $\mathbb{R}^*$?
I would like to know if I missed anything crucial. Thanks!
Best Answer
You've distorted the definition of the kernel a bit. You should have $$ \ker \varphi = \{z \in \Bbb C^* \mid \varphi(z) = 1\} $$ Because every complex number can be written as $z = x + iy$ for $x,y \in \Bbb R$, we can rewrite this as $$ \ker \varphi = \{x + iy \mid x \in \Bbb R, y \in \Bbb R, x^2 + y^2 = 1\} $$ which is to say that the kernel of $\varphi$ is the unit circle in the complex plane.
As for the image, note that $\Bbb R^*$ includes the negative numbers, which do not lie in the image of $\varphi$. You can use your idea, however, to show that every positive real lies in the image of $\varphi$. In particular: if $x$ is real and positive, then $x \in \Bbb R^* \subset \Bbb C^*$ and we have $\varphi(\sqrt{x}) = x$.