Question

If z and w are complex numbers w = a+bi and z = c+di Prove algebraically that |zw| = |z||w|.

Answer 1

Answer:

To show that $\lvert zw\rvert =\lvert z\rvert \lvert w\rvert$ observe that:

Step-by-step explanation:

We have that $w=(a+bi), z=(c+di)$. Then $zw=(c+di)(a+bi)=(ac-bd)+(cb+ad)i$. Now, the absolute value of this product is $\lvert zw \rvert =\sqrt{(ac-bd)^{2}+(cb+ad)^{2}}=\sqrt{a^{2}c^{2}-2acbd+b^{2}d^{2}+c^{2}b^{2}+2abcd+a^{2}d^{2}}=\sqrt{a^{2}c^{2}+b^{2}d^{2}+c^{2}b^{2}+a^{2}d^{2}}.$

On the other hand, $\lvert z\rvert =\sqrt{c^{2}+d^{2}}$ and $\lvert w \rvert =\sqrt{a^{2}+b^{2}}.$ Then,

$\lvert z\rvert \lvert w \rvert =\sqrt{c^{2}+b^{2}}\sqrt{a^{2}+b^{2}}=\sqrt{(c^{2}+d^{2})(a^{2}+b^{2})}=\sqrt{c^{2}a^{2}+c^{2}b^{2}+d^{2}a^{2}+d^{2}b^{2}}$

Then, it is clear that  $\lvert zw \rvert =\lvert z\rvert \lvert w \rvert$.