Complex Modulus of Product of Complex Numbers/General Result

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Theorem

Let $z_1, z_2, \ldots, z_n \in \C$ be complex numbers.

Let $\cmod z$ be the modulus of $z$.


Then:

$\cmod {z_1 z_2 \cdots z_n} = \cmod {z_1} \cdot \cmod {z_2} \cdots \cmod {z_n}$


Proof

Proof by induction:

For all $n \in \N_{> 0}$, let $P \left({n}\right)$ be the proposition:

$\cmod {z_1 z_2 \cdots z_n} = \cmod {z_1} \cdot \cmod {z_2} \cdots \cmod {z_n}$


$P \left({1}\right)$ is trivially true:

$\cmod {z_1} = \cmod {z_1}$


Basis for the Induction

$P \left({2}\right)$ is the case:

$\cmod {z_1 z_2} = \cmod {z_1} \cdot \cmod {z_2}$

which has been proved in Complex Modulus of Product of Complex Numbers.

This is our basis for the induction.


Induction Hypothesis

Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true.


So this is our induction hypothesis:

$\cmod {z_1 z_2 \cdots z_k} = \cmod {z_1} \cdot \cmod {z_2} \cdots \cmod {z_k}$


Then we need to show:

$\cmod {z_1 z_2 \cdots z_{k + 1} } = \cmod {z_1} \cdot \cmod {z_2} \cdots \cmod {z_{k + 1} }$


Induction Step

This is our induction step:


\(\ds \cmod {z_1 z_2 \cdots z_{k + 1} }\) \(=\) \(\ds \cmod {\left({z_1 z_2 \cdots z_k}\right) z_{k + 1} }\)
\(\ds \) \(=\) \(\ds \cmod {z_1 z_2 \cdots z_k} \cdot \cmod {z_{k + 1} }\) Basis for the Induction
\(\ds \) \(=\) \(\ds \cmod {z_1} \cdot \cmod {z_2} \cdots \cmod {z_k} \cdot \cmod {z_{k + 1} }\) Induction Hypothesis


So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction.


Therefore:

$\forall n \in \N_{>0}: \cmod {z_1 z_2 \cdots z_n} = \cmod {z_1} \cdot \cmod {z_2} \cdots \cmod {z_n}$

$\blacksquare$


Sources

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