Product of Complex Conjugates/General Result

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Theorem

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

Let $\overline z$ be the complex conjugate of the complex number $z$.


Then:

$\ds \overline {\prod_{j \mathop = 1}^n z_j} = \prod_{j \mathop = 1}^n \overline {z_j}$


That is: the conjugate of the product equals the product of the conjugates.


Proof

Proof by induction:

For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:

$\ds \overline {\prod_{j \mathop = 1}^n z_j} = \prod_{j \mathop = 1}^n \overline {z_j}$


$\map P 1$ is trivially true, as this just says $\overline {z_1} = \overline {z_1}$.


Basis for the Induction

$\map P 2$ is the case:

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

which has been proved in Product of Complex Conjugates.

This is our basis for the induction.


Induction Hypothesis

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


So this is our induction hypothesis:

$\ds \overline {\prod_{j \mathop = 1}^k z_j} = \prod_{j \mathop = 1}^k \overline {z_j}$

Then we need to show:

$\ds \overline {\prod_{j \mathop = 1}^{k + 1} z_j} = \prod_{j \mathop = 1}^{k + 1} \overline {z_j}$


Induction Step

This is our induction step:


\(\ds \overline {\prod_{j \mathop = 1}^{k + 1} z_j}\) \(=\) \(\ds \overline {\paren {\prod_{j \mathop = 1}^k z_j} z_{k + 1} }\)
\(\ds \) \(=\) \(\ds \overline {\paren {\prod_{j \mathop = 1}^k z_j} } \cdot \overline {z_{k + 1} }\) Basis for the Induction
\(\ds \) \(=\) \(\ds \paren {\prod_{j \mathop = 1}^k \overline {z_j} } \cdot \overline {z_{k + 1} }\) Induction Hypothesis
\(\ds \) \(=\) \(\ds \prod_{j \mathop = 1}^{k + 1} \overline {z_j}\)

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


Therefore:

$\ds \forall n \in \N: \overline {\prod_{j \mathop = 1}^n z_j} = \prod_{j \mathop = 1}^n \overline {z_j}$

$\blacksquare$


Sources

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