Complex Modulus of Product of Complex Numbers/General Result
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
- 1981: Murray R. Spiegel: Theory and Problems of Complex Variables (SI ed.) ... (previous) ... (next): $1$: Complex Numbers: Absolute Value: $1$