Proof. Let $z_{k}=x_{k}+y_{k} i$ for $k=1,2$. Then

$$

z_{1} z_{2}=\left(x_{1}+y_{1} i\right)\left(x_{2}+y_{2} i\right)=\left(x_{1} x_{2}-y_{1} y_{2}\right)+i\left(x_{2} y_{1}+x_{1} y_{2}\right)

$$

and hence

$$

f\left(z_{1} z_{2}\right)=\left[\begin{array}{cc}

x_{1} x_{2}-y_{1} y_{2} & x_{2} y_{1}+x_{1} y_{2} \\

-x_{2} y_{1}-x_{1} y_{2} & x_{1} x_{2}-y_{1} y_{2}

\end{array}\right]

$$

On the other hand,

$$

f\left(z_{1}\right) f\left(z_{2}\right)=\left[\begin{array}{cc}

x_{1} & y_{1} \\

-y_{1} & x_{1}

\end{array}\right]\left[\begin{array}{cc}

x_{2} & y_{2} \\

-y_{2} & x_{2}

\end{array}\right]=\left[\begin{array}{cc}

x_{1} x_{2}-y_{1} y_{2} & x_{2} y_{1}+x_{1} y_{2} \\

-x_{2} y_{1}-x_{1} y_{2} & x_{1} x_{2}-y_{1} y_{2}

\end{array}\right] .

$$

Therefore, $f\left(z_{1} z_{2}\right)=f\left(z_{1}\right) f\left(z_{2}\right)$