Power of Complex Modulus equals Complex Modulus of Power

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $z \in \C$ be a complex number.

Let $\left\vert{z}\right\vert$ be the modulus of $z$.

Let $n \in \Z_{\ge 0}$ be a positive integer.


Then:

$\left\vert{z^n}\right\vert = \left\vert{z}\right\vert^n$


Proof

The proof proceeds by induction.

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

$\left\vert{z^n}\right\vert = \left\vert{z}\right\vert^n$


$P \left({0}\right)$ is the case:

\(\ds \left\vert{z^0}\right\vert\) \(=\) \(\ds \left\vert{1}\right\vert\)
\(\ds \) \(=\) \(\ds 1\)
\(\ds \) \(=\) \(\ds \left\vert{z}\right\vert^0\)

Thus $P \left({0}\right)$ is seen to hold.


Basis for the Induction

$P \left({1}\right)$ is the case:

\(\ds \left\vert{z^1}\right\vert\) \(=\) \(\ds \left\vert{z}\right\vert\)
\(\ds \) \(=\) \(\ds \left\vert{z}\right\vert^1\)


Thus $P \left({1}\right)$ is seen to hold.


This is the basis for the induction.


Basis for the Induction

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

$\left\vert{z^2}\right\vert = \left\vert{z}\right\vert^2$

which is demonstrated in Square of Complex Modulus equals Complex Modulus of Square.


Thus $P \left({2}\right)$ is seen to hold.


This is the basis for the induction.


Induction Hypothesis

Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k + 1}\right)$ is true.


So this is the induction hypothesis:

$\left\vert{z^k}\right\vert = \left\vert{z}\right\vert^k$


from which it is to be shown that:

$\left\vert{z^{k + 1} }\right\vert = \left\vert{z}\right\vert^{k + 1}$


Induction Step

This is the induction step:

\(\ds \left\vert{z^{k + 1} }\right\vert\) \(=\) \(\ds \left\vert{z^k \cdot z}\right\vert\)
\(\ds \) \(=\) \(\ds \left\vert{z^k}\right\vert \cdot \left\vert{z}\right\vert\) Complex Modulus of Product of Complex Numbers
\(\ds \) \(=\) \(\ds \left\vert{z}\right\vert^k \cdot \left\vert{z}\right\vert\) Induction Hypothesis
\(\ds \) \(=\) \(\ds \left\vert{z}\right\vert^{k + 1}\)

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


Therefore:

$\forall n \in \ Z_{\ge 0}: \left\vert{z^n}\right\vert = \left\vert{z}\right\vert^n$

$\blacksquare$


Examples

Complex Modulus of $\paren {1 + i}^4$

$\left\vert{\left({1 + i}\right)^4}\right\rvert = 4$


Complex Modulus of $\paren {2 - i}^6$

$\cmod {\paren {2 - i}^6} = 125$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Power_of_Complex_Modulus_equals_Complex_Modulus_of_Power&oldid=362051"