Show that $(K,\circ)$ is a group. Let $K$ be the set of functions defined by : $f :  \mathbb C \times \mathbb C \to \mathbb C \times \mathbb C$, such that $\exists a \in \mathbb C, \exists b \in \mathbb C$, $a,b$ not simultaneously equal to zero with : $f(u,v) = (au+bv,-\bar b u+\bar a v$)
Q:Show that $(K,\circ)$ is a group.
I was able to show  $f_1 \circ f_2 \in K$ (Closure) and $(f_1 \circ f_2)\circ f_3 = f_1 \circ (f_2 \circ f_3)$ (Associativity) . but i'm not familiar nor able to show the symmetry nor the identity element of this set.
I'd appreciate any help i can get that would push me in the right direction! thanks. 
 A: According to you, you have already shown the closure of such functions. Also, associativity is always true for composition of functions, no matter what. To prove that $K$ is a group under composition of functions, we have to show the existence of an identity element in the group, and the existence of the inverse of any element in the group.
The identity function will be the identity element of $K$ and it is in $K$ because for $a=1,b=0$, we have $f(u,v)=(u,v)$. Therefore, the identity function is in the set $K$.
Now to find the inverse of $f$, you have to solve the following system of equations in $\mathbb{C}^2$,
$$au+bv=c$$ $$-\bar{b}u+\bar{a}v=d$$
You want to solve it for the unknown $u$ and $v$. 
The determinant is $|a|^2+|b|^2$ which is not $0$ because $a,b$ are not simultaneously $0$. Therefore, the system has a unique solution. Let's find it:
$$\pmatrix{u \\ v} = \frac{1}{|a|^2+|b|^2}\pmatrix{\bar{a} && -b \\ \bar{b}  && a}\pmatrix{c \\d}$$
$$\pmatrix{u \\ v} = \pmatrix{ \frac{\bar{a}c-bd}{|a|^2+|b|^2} \\ \frac{\bar{b}c+ad}{|a|^2+|b|^2}}$$
If you set $$r=\frac{\bar{a}}{|a|^2+|b|^2}$$ $$s=\frac{-b}{{|a|^2+|b|^2}}$$
You obtain, $$f^{-1}(c,d)=(rc+sd,-\bar{r}c+\bar{s}d)$$ which is of the given form and therefore, belongs to $K$.
