Show $(M,+)$ is isomorphic to $(\mathbb{C},+)$ 
Let, $M=\{ \begin{pmatrix}a&-b\\b&a\end{pmatrix} :a,b\in \mathbb{R}\}$, show $(H,+) $ is isomorphic as a binary structure to $(\mathbb{C},+)$

Define $f : M\rightarrow \mathbb{C} $ by $f\begin{pmatrix}a&-b\\b&a\end{pmatrix} = a+bi$
Let $a,b,c,d\in \mathbb{R}$
$1-1$:
Suppose  $f\begin{pmatrix}a&-b\\b&a\end{pmatrix}=f\begin{pmatrix}c&-d\\d&c\end{pmatrix}$, then $a+bi =c+di$, thus $a=c$ and $b=d$, so $f$ is one to one.
Onto:
Let $a+bi\in \mathbb{C}$ , then $\begin{pmatrix}a&-b\\b&a\end{pmatrix}\in M$, so $f\begin{pmatrix}a&-b\\b&a\end{pmatrix}=a+bi$, thus $f$ is onto.
Homomorphic:
$\begin{align*} &f(\begin{pmatrix}a&-b\\b&a\end{pmatrix} + \begin{pmatrix}c&-d\\d&c\end{pmatrix})\\
=&f(\begin{pmatrix}a+c&-(b+d)\\b+d&a+c\end{pmatrix})\\
= &(a+c)+(b+d)i\\
= &f(\begin{pmatrix}a&-b\\b&a\end{pmatrix}) +f(\begin{pmatrix}c&-d\\d&c\end{pmatrix})\end{align*}$
I don't think I show $f: M\rightarrow \mathbb{C} $ is $1-1$ and onto correctly.
can any give me a hit to show $f: M\rightarrow \mathbb{C}$ is $1-1 $ and onto? thanks!

Responding to the request from @Julian and to move this question to answered question list, I will try to sum up what has been mentioned in the comments.
Yes, your answer is correct.  
