Proving A is a group over unusual operation Let $A= R^2 - {(0,0)}$ and operation $*$ over A is defined by.
$$(a, b)*(a′, b′) = (aa′−bb′, ab′+a′b)$$
Question: Is $(A,*)$ a group ?
My attempt: First of all, as for all $a$,$b$ in $A$ $$(1,0)*(a,b)=(a.1-0.b,b.1-a.0)=(a,b)=(a,b)*(1,0)$$
$(1,0)$ is identity element.
Secondly, let $(a,b)$ and $(c,d)$ be elements of $A$, Then:
$$(a,b)*(c,d)=(ac-bd,ad+cb)$$
As both $ac-bd$ and $ab+cb$ in $R^2$, $A$ is closed over $*$.
I had problem with proving the existence of inverse for every element of A.

Note: if we identify $\mathbb{R}^2$ with $\mathbb{C}$ via $(a,b)\equiv a+ib$, then $(a,b)*(c,d)=(ac-bd, ad+cb)\equiv(ac-bd)+i(ad+bc)=(a+ib)\cdot(c+id)$. So what you are checking is actually whether $(\mathbb{C}-\{0\},\cdot)$ where $\cdot$ denotes the usual multiplication of complex numbers is a group. But even cats and dogs will say that this is true when asked.
