If z and w are complex numbers w = a+bi and z = c+diProve algebraically that |zw| = |z||w|.
Accepted Solution
A:
Answer:To show that [tex]\lvert zw\rvert =\lvert z\rvert \lvert w\rvert [/tex] observe that:Step-by-step explanation:We have that [tex]w=(a+bi), z=(c+di)[/tex]. Then [tex]zw=(c+di)(a+bi)=(ac-bd)+(cb+ad)i[/tex]. Now, the absolute value of this product is [tex]\lvert zw \rvert =\sqrt{(ac-bd)^{2}+(cb+ad)^{2}}=\sqrt{a^{2}c^{2}-2acbd+b^{2}d^{2}+c^{2}b^{2}+2abcd+a^{2}d^{2}}=\sqrt{a^{2}c^{2}+b^{2}d^{2}+c^{2}b^{2}+a^{2}d^{2}}.[/tex]On the other hand, [tex]\lvert z\rvert =\sqrt{c^{2}+d^{2}}[/tex] and [tex]\lvert w \rvert =\sqrt{a^{2}+b^{2}}.[/tex] Then, [tex]\lvert z\rvert \lvert w \rvert =\sqrt{c^{2}+b^{2}}\sqrt{a^{2}+b^{2}}=\sqrt{(c^{2}+d^{2})(a^{2}+b^{2})}=\sqrt{c^{2}a^{2}+c^{2}b^{2}+d^{2}a^{2}+d^{2}b^{2}}[/tex]Then, it is clear that Β [tex]\lvert zw \rvert =\lvert z\rvert \lvert w \rvert[/tex].