Zeige:
\(\left| {{z_1} \cdot {z_2}} \right| = \left| {{z_1}} \right| \cdot \left| {{z_2}} \right|\)
\(\sqrt {a_1^2a_2^2 + b_1^2b_2^2 + a_1^2b_2^2 + a_2^2b_1^2} = \sqrt {a_1^2a_2^2 + b_1^2b_2^2 + a_1^2b_2^2 + a_2^2b_1^2} \)
\({z_1}^2 \cdot {z_2}^2 = {z_1}^2 \cdot {z_2}^2\)
\({z_1} \cdot {z_2} = {z_1} \cdot {z_2}\)
\(\sqrt {{z_1}^2 \cdot {z_2}^2} = \sqrt {{z_1}^2 \cdot {z_2}^2} \)
Ich errechne eine abweichende Lösung