If $z$ and $w$ are complex numbers can we use the proof in $\mathbb{R}$ to demonstrate that $|z w|=|z||w|$?

Question

If yes could you explain why? Sorry if the question is trivial but I'm new to complex numbers and I see lots of examples where properties of real numbers are used in complex without to prove it . This really surprise me since complex numbers are a superset of real numbers. For example I saw in my book $z*1=z$ in $ \mathbb{C} $ I had to make the complex multiplication to be sure that it was true, because this 1 is in $\mathbb{C} $ (1,0)

Answer 1

First note that, for $z=a+ib$ we have $z\overline{z}=\sqrt{a^2+b^2}=|z|^2$ and also $\overline{z_1z_2}=\overline{z_1}\,\overline{z_2}.$

Therefore, $|zw|^2=(zw)\overline{zw}=zw\overline{z}\,\overline{w}=z\overline{z}w\overline{w}=|z|^2|w|^2.$ This implies $|zw|=|z||w|.$

Answer 2

Let $z=a+bi$ and $w=c+di$ where $(a,b,c,d)\in\mathbb{R}^{4}$. The modulus $|z|$ is defined by $|z|=\sqrt{a^{2}+b^{2}}$ and $|w|=\sqrt{c^{2}+d^{2}}$, thus $$\begin{align} |z||w| &=\sqrt{a^{2}+b^{2}}\sqrt{c^{2}+d^{2}}\\ &=\sqrt{a^{2}c^{2}+a^{2}d^{2}+b^{2}c^{2}+b^{2}d^{2}} \end{align}$$

Now, consider $zw=(a+bi)(c+di)=(ac-bd)+(ad+bc)i$. We have

$$\begin{align} |zw|&=\sqrt{(ac-bd)^{2}+(ad+bc)^{2}}\\ &=\sqrt{a^{2}c^{2}+b^{2}d^{2}-2abcd+a^{2}d^{2}+b^{2}c^{2}+2abcd}\\ &=\sqrt{a^{2}c^{2}+b^{2}d^{2}+a^{2}d^{2}+b^{2}c^{2}}\\ &=|z||w| \end{align}$$