I am self studying from A Book of Abstract Algebra by Charles C. Pinter. In chapter 3 problem set B problem 4 it asks to show whether $(a,b)\star(c,d)=(ac-bd,ad+bc)$ on the set $ \{(x,y) \in \mathbb R^2 | (x,y) \not= (0,0)\} $ is a group. Currently I am in the middle of showing that there is an inverse. This is the equation I am trying to solve to show that there is an inverse.
$(a,b) \star (a',b')=(1,0)=(aa' - bb',ab' + ba')$
I am pretty sure I have the answer for the first entry of the inverse element but I pass through steps with division. In particular I am wondering whether $[1]$ to $[2]$ is a valid step and if it is why. In other words if I exclude $0$ at one step do I have to keep $0$ excluded for the rest of the proof? If its not legal would I just have to split the problem into a couple of cases?
$[1]$ $\frac{aa'-1}{b} =\frac{-ba'}{a}$ where $a\not=0$ and $b\not=0\\$
$[2]$ $a^2 a' - a = -b^2 a'$ (note I did not exclude $0$)