Λύση

Έστω $x_1+y_1i, x_2+y_2i\in \mathbb C$ και $c\in \mathbb R$ τότε

$$F((x_1+y_1i)+(x_2+y_2i))=F((x_1+x_2)+(y_1+y_2)i)= \left(\beg... ... x_1+x_2 & y_1+y_2\\ -y_1-y_2 & x_1+x_2 \end{array}\right)=$$

$$\left(\begin{array}{cc} x_1 & y_1\\ -y_1 & x_1 \end{array... ...y_2\\ -y_2 & x_2 \end{array}\right)=F(x_1+y_1i)+F(x_2+y_2i).$$

Επίσης,

$$F(c(x_1+y_1i))=F(cx_1+cy_1i) \left(\begin{array}{cc} cx_1 & cy_1\\ -cy_1 & cx_1 \end{array}\right)=$$

$$c\left(\begin{array}{cc} x_1 & y_1\\ -y_1 & x_1 \end{array}\right)=cF(x_1+y_1i).$$

Εύκολα βλέπουμε ότι ${\rm Ker}F={\bf0}$ και άρα η $F$ είναι 1-1. Τέλος,

$$F(z_1z_2)=F((x_1+y_1i)(x_2+y_2i))=F((x_1x_2-y_1y_2)+(x_1y_2+x_2y_1)i)=$$

$$\left(\begin{array}{cc} x_1x_2-y_1y_2 & x_1y_2+x_2y_1\\ -x_... ...begin{array}{cc} x_2 & y_2\\ -y_2 & x_2 \end{array}\right)=$$

$$F(x_1+y_1i)F(x_2+y_2i)=F(z_1)F(z_2).$$