Question: I have a mapping $$\theta:S\to \mathbb C $$ where $\mathbb C$ is the field of complex numbers and $$S=\left[\begin{array}{cc} a & -b \\b & a\end{array}\right]$$ The mapping is defined by $$\theta\Bigg(\left[\begin{array}{cc} a & -b \\b & a\end{array}\right] \Bigg)=a+ib$$

where a and b belong to the real numbers.

How do I show injection and surjection?

Answer: 
To show $\phi$ is injective, suppose $\phi( \left[\begin{array}{cc} a & -b \\b & a\end{array}\right])=\phi( \left[\begin{array}{cc} c & -d \\d & c\end{array}\right])$. That is, $a+ib=c+id$, which is true if and only if $a=c$ and $b=d$ Hence, $\left[\begin{array}{cc} a & -b \\b & a\end{array}\right]= \left[\begin{array}{cc} c & -d \\d & c\end{array}\right]$   

To show $\phi$ is surjective, let $x+iy \in \mathbb{C}$. Now $\left[\begin{array}{cc} x & -y \\y & x\end{array}\right] \in S$ and $\phi(\left[\begin{array}{cc} x & -y \\y & x\end{array}\right])=x+iy$
Injectivity:

Suppose $$\phi\left(\begin{pmatrix} a & -b \\ b & a \end{pmatrix}\right) = 0 + 0i$$ What can you say about $\begin{pmatrix} a & -b \\ b & a \end{pmatrix}$? Is this sufficient (i.e. why do I only need to look at the pre-image of $0 + 0i$?)

Surjectivity:

Choose any $z = x + iy \in \Bbb C$. Can you find $\begin{pmatrix} a & -b \\ b & a \end{pmatrix}$ so that

$$
\phi\left(\begin{pmatrix} a & -b \\ b & a \end{pmatrix}\right) = x + yi?
$$

(In other words, how can you choose $a,b$ so this happens?)