Power of Complex Conjugate is Complex Conjugate of Power

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

Let $\overline z$ denote the complex conjugate of $z$.

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

Then:
 * $\overline {z^n} = \left({\overline z}\right)^n$

Proof
The proof proceeds by induction.

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

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

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

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

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:


 * $\overline {z^2} = \left({\overline z}\right)^2$

which is demonstrated in Square of Complex Conjugate is Complex Conjugate 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:
 * $\overline {z^k} = \left({\overline z}\right)^k$

from which it is to be shown that:
 * $\overline {z^{k + 1} } = \left({\overline z}\right)^{k + 1}$

Induction Step
This is the induction step:

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

Therefore:
 * $\forall n \in \ Z_{\ge 0}: \overline {z^n} = \left({\overline z}\right)^n$