Why can we distribute the complex modulus? (looking for intuition/proof)

Question

if you have,

$$ z= z_1 \cdot z_2 ... z_n$$

then,

$$ |z| = |z_1 \cdot z_2 ... z_n| = |z_1| |z_2| |z_3| ... |z_n|$$

Now, why is this equality true? $$|z_1 \cdot z_2 ... z_n| = |z_1| |z_2| |z_3| ... |z_n|$$

Answer 1

Think about the polar form of a complex number.

If each $\displaystyle z_j=r_je^{i\theta_j}$, where $r_j=|z_j|$ (the modulus), and $\theta_j$ is the argument.

Then $z=z_1z_2z_3....z_n=(r_1r_2r_3...r_n)e^{i(\theta_1+\theta_2+\theta_3...\theta_n)}$

Answer 2

A gory proof(as I'm unable to provide intuition): It is enough to prove it for two: $z=x\cdot y$. This can be inductively extended for $n$.

Let $x=a+ib, y=c+id\Rightarrow z=(ac-bd)+i(ad+bc)$. We now have $$|z|=\sqrt{(ac-bd)^2+(ad+bc)^2}$$ And $$|x|=\sqrt{a^2+b^2},|y|=\sqrt{c^2+d^2}$$ This implies $$|x\cdot y|=\sqrt{(a^2+b^2)(c^2+d^2)}$$ Now of you compare, you shall find that the statement is indeed true.

Answer 3

$$ z_1 = r_1e^{i\theta}$$ $$z_2 = r_2 e^{i\phi}$$ $$ z_1 * z_2 = r_1 * r_2 e^{i \theta + i \phi}$$ $$ |z_1 z_2| = r_1 r_2$$ $$|z_1| = r_1$$ $$ |z_2| = r_2$$ $$ |z_1| * |z_2| = r_1 * r_2$$

Now, let P(n) contain the statement that $ |z_1| * |z_2| ... |z_n| = | z_1 z_2 z_3..z_n|$

$$P(1) = |z_1| = |z_1|$$

$$P(k) = |z_1| |z_2| |z_3|.......|z_k| = |z_1 z_2 z_3...z_k|$$

$$ P(k+1) = |z_1| |z_2| |z_3|.......|z_k||z_{k+1}| = |z_1 z_2 z_3...z_k| |z_{k+1}| =|z_1 z_2 z_3...z_k z_{k+1}| $$

q.e.d