Show that the system of all matrices of the special form
$$ \begin{pmatrix} \alpha & \beta \\ -\beta & \alpha \\ \end{pmatrix} $$
combined by matrix addition and matrix multiplication, is isomorphic to the field of complex number.
How do i start showing this ?
And first of all how can matrix have a one to one connection with complex numbers
An isomorphism is given by $$\begin{bmatrix}\alpha&\beta\\-\beta&\alpha\end{bmatrix}\mapsto\alpha+\beta i.$$ Another isomorphism is given by $$\begin{bmatrix}\alpha&\beta\\-\beta&\alpha\end{bmatrix}\mapsto\alpha-\beta i.$$ You just need to show that one of the maps really does preserve addition, multiplication and multiplicative identity.
(Edit: You may want to ask how I come up with these two maps, or why do we even consider the matrices of such form in the first place. If you want, I can add details.)
(Edit 2: You need to show the map is bijective as well.)
Added: To show it preserves multiplication, you are trying the prove the equality $$F\left(\begin{bmatrix}\alpha&\beta\\-\beta&\alpha \end{bmatrix}\begin{bmatrix}\gamma&\sigma\\-\sigma&\gamma \end{bmatrix}\right)=F\left(\begin{bmatrix}\alpha&\beta\\-\beta&\alpha \end{bmatrix}\right)F\left(\begin{bmatrix}\gamma&\sigma\\-\sigma&\gamma \end{bmatrix}\right)$$ holds, where $$F\left(\begin{bmatrix}\alpha&\beta\\-\beta&\alpha \end{bmatrix}\right)=\alpha+\beta i.$$ It is just a slightly tedious calculation: $$F\left(\begin{bmatrix}\alpha&\beta\\-\beta&\alpha \end{bmatrix}\begin{bmatrix}\gamma&\sigma\\-\sigma&\gamma \end{bmatrix}\right)=F\left(\begin{bmatrix}\alpha\gamma-\beta\sigma&\alpha\sigma+\beta\gamma\\-(\beta\gamma+\alpha\sigma)&-\beta\sigma+\alpha\gamma\end{bmatrix}\right)=\color{red}{(\alpha\gamma-\beta\sigma)+(\alpha\sigma+\beta\gamma)i},\\ F\left(\begin{bmatrix}\alpha&\beta\\-\beta&\alpha \end{bmatrix}\right)F\left(\begin{bmatrix}\gamma&\sigma\\-\sigma&\gamma \end{bmatrix}\right)=(\alpha+\beta i)(\gamma+\sigma i)=\color{red}{(\alpha\gamma-\beta\sigma)+(\alpha\sigma+\beta\gamma)i}.$$
As for how we came up with the isomorphisms, we may do this algebraically:
Multiplicative identity must be mapped to multiplicative identity. In this case, the identity matrix must be mapped to $1+0i$.
The complex polynomial equation $z^2+1=0$ has complex number solutions $i$ and $-i$. If there is isomorphism from the set of matrices to complex numbers, it means the matrix polynomial equation $X^2+I=\boldsymbol0$ has two solutions $$\begin{bmatrix}\alpha&\beta\\-\beta&\alpha\end{bmatrix},\ \begin{bmatrix}-\alpha&-\beta\\\beta&-\alpha\end{bmatrix}$$ where $I$ is the identity matrix, $\boldsymbol0$ is the zero matrix, and the two solutions shall be mapped to $i$ and $-i$ by an isomorphism. When you solve the matrix polynomial equation, you would find out the two unknowns $\alpha,\beta$ has to satisfy $$\begin{cases}\alpha^2-\beta^2=-1\\2\alpha\beta=0\end{cases}.$$ This would mean $\beta$ cannot be zero ($\alpha^2=-1$ cannot hold for a real number $\alpha$), $\alpha$ must be zero, and $\beta=1$ or $\beta=-1$.
The facts that isomorphism preserves real scalar multiplication (if you do consider scalar multiplication), and that $\begin{bmatrix}1&0\\0&1\end{bmatrix}$ is sent to $1$, $\begin{bmatrix}0&1\\-1&0\end{bmatrix}$ is sent to $i$ or $-i$, force us to define $$\begin{bmatrix}\alpha&\beta\\-\beta&\alpha\end{bmatrix}\mapsto\alpha+\beta i$$ or $$\begin{bmatrix}\alpha&\beta\\-\beta&\alpha\end{bmatrix}\mapsto\alpha-\beta i.$$
As for why we came up with these matrices of special form, it follows from a fact about complex number multiplication: multiplication by a complex number is the same as doing a proper rotation and then a scaling. The matrix $$\begin{bmatrix}\alpha&\beta\\-\beta&\alpha\end{bmatrix}$$ represents a linear transformation that does a rotation and then a scaling. You may notice that the matrix may be written as $$\begin{bmatrix}r\cos\theta&r\sin\theta\\-r\sin\theta&r\cos\theta\end{bmatrix},$$ where $r$ is the scaling factor and $\theta$ is the rotation angle. This is a whole new story I won't extend further.
It will be sum of kronecker product of
$\begin{bmatrix}1&0\\0&1\end{bmatrix}$ and $\alpha$
$\begin{bmatrix}0&1\\-1&0\end{bmatrix}$ and $\beta$
By block multiplication property of matrices:
1) should be obvious.
We can verify 2) by calculating it's first few powers up to 4 to see it shifts through $-1,-i,1,\cdots$ corresponding to $\begin{bmatrix}-1&0\\0&-1\end{bmatrix} ,\begin{bmatrix}0&-1\\1&0\end{bmatrix}, \begin{bmatrix}1&0\\0&1\end{bmatrix}$ above.
