Define multiplication in $\mathbb{R} \times \mathbb{R}$ by $(a,b)(c,d)=(ac,bd)$ and $f$ by $f(a+bi)=(a,b)$. To show $f$ is an isomorphism, it would suffice to show that $f$ is a bijection and a homomorphism.
I am easily able to show that f is a bijection. However, showing that f is a homomorphism is giving me some trouble. I can see that
$f((a+bi)+(c+di))=f((a+c)+(b+d)i)=(a+c,b+d)=(a,b)+(c,d)=f(a + bi)+f(c+di)$
however,
$f((a+bi)(c+di))=f(ac+adi+bci-bd)=f((ac-bd)+(ad+bc))=(ac-bd,ad+bc) \neq (ac,bd)$.
Where am I going wrong here?
Also, another thing I've noticed - If $f$ is an isomorphism, then the multiplicative identity in $\mathbb{C}$ should be mapped to the multiplicative identity in $\mathbb{R} \times \mathbb{R}$. But $1+0i=1$ is the multiplicative identity in $\mathbb{C}$ and $f(1)=(1,0)$. This isn't the multiplicative identity in $\mathbb{R} \times \mathbb{R}$ because $(1,1)$ is!
How can $f$ possibly be an isomorphism?
I'm sure my logic is wrong in at least one place. Any critiques are appreciated.