Prove this function is a group homomorphism?

Question

Let $G = (\mathbb{R}-{0}, \times) \oplus (\mathbb{R}, +)$ and let the function $f:G \rightarrow (\mathbb{C}-0, \times)$ be defined by $f(r,\theta) = (r\cos\theta) + (r\sin\theta)i$.

I am trying to prove that $f$ is a group homomorphism. I'm doing this by showing that for any two pairs/elements $(a,b),(c,d)$ in G, that $f(a+c,b+d)=f(a,b)f(c,d)$.

I've already let $(a,b),(c,d) \in G$, and so $f(ac,b+d)=(ac)(\cos(b+d)) + (ac)(\sin(b+d))i$.

$=(ac)(\cos(b+d) + \sin(b+d)i)$ $=(ac)((\cos b\cos d - \sin b \sin d) + (\sin b \cos d + \cos b \sin d)i)$ $=(ac)((\cos b\cos d - i\sin b \cos d + i\cos b \sin d- \sin b \sin d) )$ $=(ac)((\cos d (\cos b - i\sin b) + \sin d(i\cos b - \sin b) )$ ...

I've also started going the other way from what $f(a,b)f(c,d)$ should equal but it seems to be almost neverending rabbit trails using sine and cosine properties. Any advice?

Answer 1

You want to show that $f(ac, b+d)=f(a,b)f(c,d)$. The operation in the first factor of $G$ is the multiplication of real numbers. That is why $a$ and $c$ are being multiplied.

The LHS is $ac\cos(b+d)+iac\sin(b+d)$.

The RHS is $(a\cos(b)+ia\sin(b))(c\cos(d)+ic\sin(d))$

Use that $\sin(x+y)=\sin(x)\cos(y)+\cos(x)\sin(y)$ and that $\cos(x+y)=\cos(x)\sin(y)-\sin(x)\cos(y)$ and you get it.

Answer 2

The group operation in ${\bf{R}}^{\times}\times{\bf{R}}$ is $(r,\alpha)(s,\beta)=(rs,\alpha+\beta)$ and the group operation in ${\bf{C}}^{\times}$ is the complex multiplication of complex numbers.

So all you need to check is if $f((r,\alpha)(s,\beta))=f(rs,\alpha+\beta)=f(r,\alpha)\cdot f(s,\beta)$.