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Let $f$ be the map sending each complex number
$$z=x+y i \rightarrow\left[\begin{array}{cc} x & y \\ -y & x \end{array}\right]$$

Show that $f\left(z_{1} z_{2}\right)=f\left(z_{1}\right) f\left(z_{2}\right)$ for all $z_{1}, z_{2} \in \mathbb{C}$.
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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, $f\left(z_{1} z_{2}\right)=f\left(z_{1}\right) f\left(z_{2}\right)$
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