Proofs with steps of division and setting things not equal to zero 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 $0$)
 A: You want $a'$ and $b'$ such that $aa'-bb'=1$ and $ab'+ba'=0$. In effect you want to show that the system 
$$\left\{\begin{align*}
&ax-by=1\\
&bx+ay=0
\end{align*}\right.\tag{1}$$
always has a solution with $\langle x,y\rangle\ne\langle 0,0\rangle$ provided that $\langle a,b\rangle\ne\langle 0,0\rangle$. You can do this in several ways. If you already know the relevant linear algebra, you can note that $$\det\pmatrix{a&-b\\b&a}=a^2+b^2\ge 0\;,$$
with equality iff $a=b=0$, so $(1)$ always has a unique solution, which clearly (by virtue of the first equation of $(1)$) cannot be $x=y=0$.
Alternatively, you can work out the solution by whatever technique appeals to you and then show directly that it exists and is a non-$\langle 0,0\rangle$ solution whenever $a\ne 0\ne b$.
Your approach, however, does require you to consider separate cases, since it’s entirely possible that one side of $[1]$ is undefined.
A: Hint: Let $a=r\cos\theta$, $b=r\sin\theta$, and let $a'=r'\cos\theta'$, $b'=r'\sin\theta'$. (As usual choose $r$, $r'$ positive.) It is easy to see that $r'=1/r$.  Now you can identify $\theta'$ by recalling some trigonometric identities. 
Or else if you are familiar with complex numbers, there is a much more elegant version of the same thing. We have $(a+bi)(c+di)=ac-bd++(ad+bc)i$. So 
$$\frac{1}{a+bi}=\frac{a-bi}{(a+bi)(a-bi)}=\frac{a}{a^2+b^2}-\frac{b}{a^2+b^2}i.$$
