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:
| \(\ds \overline {z^0}\) | \(=\) | \(\ds \overline 1\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 1\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \left({\overline z}\right)^0\) |
Thus $P \left({0}\right)$ is seen to hold.
Basis for the Induction
$P \left({1}\right)$ is the case:
| \(\ds \overline {z^1}\) | \(=\) | \(\ds \overline z\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \left({\overline z}\right)^1\) |
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:
- $\overline {z^2} = \left({\overline z}\right)^2$
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:
| \(\ds \overline {z^{k + 1} }\) | \(=\) | \(\ds \overline {z^k \cdot z}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \overline {z^k} \cdot \overline z\) | Product of Complex Conjugates | |||||||||||
| \(\ds \) | \(=\) | \(\ds \left({\overline z}\right)^k \cdot \overline z\) | Induction Hypothesis | |||||||||||
| \(\ds \) | \(=\) | \(\ds \left({\overline z}\right)^{k + 1}\) |
So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction.
Therefore:
- $\forall n \in \ Z_{\ge 0}: \overline {z^n} = \left({\overline z}\right)^n$
$\blacksquare$