Complex number isomorphic to certain $2\times 2$ matrices?

I have been trying to prove this, but I am having trouble understanding how to prove the following mapping I found is injective and surjective. Just as a side note, I am trying to show the complex ring is isomorphic to special $2\times2$ matrices in regard to matrix multiplication and addition. Showing these hold is simple enough.

$$\phi:a+bi \rightarrow \begin{pmatrix} a & -b \\ b & a \end{pmatrix}$$

This is what I have so far:

Injective: I am also confused over the fact that there are two operations, and in turn two neutral elements (1 and 0). Showing that the kernel is trivial is usually the way I go about proving whether a mapping is injective, but I just can't grasp this.

$$ \phi(z_1) = \phi(z_2) \implies \phi(z_1)\phi(z_2)^{-1} = I = \phi(z_1)\phi(z_2^{-1}) = \phi(z_2)\phi(z_2^{-1}).$$

So if we can just show that the kernel of $\phi$ is trivial, then it also shows that $z_1 = z_2$. The only complex number that maps to the identity matrix is one where $a = 1$ and $b = 0$, $a + bi = 1 + 0i = 1$.

Using a similar argument for addition we can just say that the only complex number $z$ such that $\phi(z) = 0\text{-matrix}$, is one where $a=0$ and $b=0$, $a+bi=0+0i=0$.

Surjective:

I forgot to add this before I posted, but I honestly don't really understand how to prove this because it just seems so obvious. All possible $2\times2$ matrices of that form have a complex representation because the complex number can always be identified by its real parts and since the elements of the $2\times2$ matrix are real then the mapping is obviously onto.

I have always had trouble understanding when I can say that I have "rigorously" proved something, so any help would be appreciated!

• For injectivity you need to show that if $\phi(z_1)=\phi(z_2)$ then $z_1=z_2$. So assume that $$\phi(a+bi)=\begin{pmatrix}a&-b\\b&a\end{pmatrix}=\begin{pmatrix}a'&-b'\\b'&a'\end{pmatrix}=\phi(a'+b'i)$$ then $$\begin{pmatrix}a-a'&-b+b'\\b-b'&a-a'\end{pmatrix}=\begin{pmatrix}0&0\\0&0\end{pmatrix}$$ so $a=a'$, and $b=b'$ which means that $a+bi=a'+b'i$.

• For surjectivity all the matrices are on the form $$\begin{pmatrix}a&-b\\b&a\end{pmatrix}$$ with $a,b\in\mathbb{R}$, and the element $a+bi$ maps to it.

You could prove injectivity simply by showing $$z\ne w \Rightarrow \phi(z) \ne \phi(w)$$ wich is next to obvious.
Regarding surjectivity: A function is always surjective on its range. All you need to show here is that any Matrix $\pmatrix{a&-b\\b&a}$ is in the range of $\phi$, explicitly $a+bi$ is the required argument (sounds like nothing to show, because it's simply the definition of $\phi$ wich guarantees this)