Showing that an operation has inverses I'm trying to show whether or not  

$(a, b) \cdot (c, d) = (ac-bd, ad+bc)$ on the set $\Bbb R \times \Bbb R$ with the origin deleted forms a group, an abelian group, or neither.

I've shown that it's commutative, associative, has an identity element $(1, 0)$ and am now thoroughly stuck on showing that it ever element in the set has an inverse. The (elementary/arithmetic) algebra just isn't working out for me here. I solved for $a^{-1}$ and $b^{-1}$ and got something rather complicated, but then when I plugged it back in to check my answer the algebra is just not working out. 
For $a^{-1}$ I got $\dfrac{1+bb^{-1}}{a}$ and for $b^{-1}$ I got $\dfrac{-b}{a^2 +b^2}$. 
I google searched the problem ("A Book of Abstract Algebra by Pinter, chapter 3 problem set B problem #4) and the question has already been asked on here but the individual there had a bit more specific of a question, and, in short I didn't find the answers helpful so please don't direct me there. For your reference however, it may be helpful to look at. It talks about complex numbers and such which this book doesn't cover so I'm wondering what the deal with that is...
So, if someone could please find the inverses and plug them back in to ensure they are true for me and give a THOROUGH explanation in doing so, it would be much appreciated! 
 A: Here, given $(a,b)$, you are simply trying to find a pair $(c,d)$ such that:
$$ (a,b)\cdot (c,d)=(1,0)$$
This corresponds exactly to the system of two equations:
$$ ac-bd=1$$
$$ bc+ad=0$$
And from this we are trying to solve for $c$ and $d$ in terms of $a$ and $b$. Luckily this is a system of two equations and two variables, and hence some simple algebra gives:
$$ c=\frac{a}{a^2+b^2}$$
$$ d=\frac{-b}{a^2+b^2}$$
Therefore the pair $\left(\frac{a}{a^2+b^2}, \frac{-b}{a^2+b^2} \right)$ is the inverse of $(a,b)$.
EDIT: Here are the details of the "simple algebra":
We multiply the first equation by $a$ and the second by $b$ to get:
$$ a^2c-abd=a$$
$$ b^2c+abd=0$$
Adding them together, we get:
$$ c(a^2+b^2)=a$$
And thus, $$ c=\frac{a}{a^2+b^2}$$
Now, from the second equation and our solution for $c$, we know:
$$ d=\frac{-bc}{a}=\frac{-ba}{a(a^2+b^2)}=\frac{-b}{a^2+b^2}$$
A: You were "close":
Instead of using the terms "$a^{-1}$" and "$b^{-1}$" let us introduce the following two functions:
$p_1: \Bbb R \times \Bbb R \to \Bbb R\\ p_2: \Bbb R \times \Bbb R \to \Bbb R$
given by:
$p_1(x,y) = x\\ p_2(x,y) = y.$
(we can call these "coordinate function number $1$", and "coordinate function number $2$").
You correctly found that $p_2((a,b)^{-1}) = \dfrac{-b}{a^2 + b^2}$, and that:
$p_1((a,b)^{-1}) = \dfrac{1 + bp_2((a,b)^{-1})}{a}$.
Substituting, we have:
$p_1((a,b)^{-1}) = \dfrac{1 + b\dfrac{-b}{a^2 + b^2}}{a} = \dfrac{\dfrac{a^2 + b^2}{a^2 + b^2}  - \dfrac{b^2}{a^2 + b^2}}{a}$
$= \dfrac{a^2}{a^2 + b^2}\cdot\dfrac{1}{a} = \dfrac{a}{a^2 + b^2}$.
Then, since $(x,y) = (p_1(x,y),p_2(x,y))$, we have:
$(a,b)^{-1} = \left(\dfrac{a}{a^2+b^2},\dfrac{-b}{a^2 + b^2}\right)$.
To simplify verifying $(a,b)(a,b)^{-1} = (1,0)$, note that:
$(a,b)(a,-b) = (a^2 - (-b^2),ab - ba) = (a^2 + b^2, 0)$
and that for any real number $r$, $(a,b)(rc,rd) = (a(rc) - b(rd), a(rd) + b(rc))$
$= (r(ac - bd), r(ad + bc))$
So letting $r = \dfrac{1}{a^2 + b^2}$, we have:
$(a,b)(ra,-rb) = (r(a^2 + b^2),r0) = (1,0)$.
A: The answers to the question "Proofs with steps of division and setting things not equal to zero" use linear algebra or complex arithmetic. Since you've already seen those derivations, I'll skip to the end and be very explicit:
$$(a,b)^{-1}=\left(\frac{a}{a^2+b^2},\frac{-b}{a^2+b^2}\right).$$
The exercise is to verify the identity
$$(a,b)\cdot\left(\frac{a}{a^2+b^2},\frac{-b}{a^2+b^2}\right)=(1,0),$$
which you should do by applying the definition of "$\cdot$" to compute the left-hand side.
