Complex conjugate to the power proof

Question

How can I proof using math induction that $$\overline{z^n} =\overline{z}^n$$ where $z$ is a complex number, $n$ is a positive whole number.

Answer 1

Start with the base case.

Suppose $n=1$. Because $z^1 = z$ for all $z \in \mathbb{C}$,

$$\overline{z^1} = \overline{z} = \overline{z}^1$$

and we're done.

Now, for our inductive step, suppose this is true for $n$. We want to show it is also true for $n+1$. But before we can do this, I claim that, for all $x,y \in \mathbb{C}$,

$$\overline{xy} = \overline{x}\overline{y}$$

Let $x = a+bi$ and $y = c+di$, where $i := \sqrt{-1}$.

$$\overline{(a+bi)(c+di)} = \overline{(ac-bd)+ (ad+bc)i} = (ac-bd)-(ad+bc)i = (a-bi)(c-di)$$

But the last part of the above is just $(\overline{a+bi})(\overline{c+di})$. Hence, we have shown our claim.

Let's get back to our inductive step. We're supposing that your claim is true for some fixed $n$, and we want to show it holds for $n+1$. To this end,

$$\overline{z^{n+1}} = \overline{z^nz} = \overline{z^n}\overline{z}$$

where the last equality holds because of our work above. Then, by our inductive hypothesis,

$$\overline{z^n} = \overline{z}^n$$

Therefore,

$$\overline{z^{n+1}} = \overline{z}^n \overline{z} = \overline{z}^{n+1}$$

and we are done.

Answer 2

Our base case is $n=1$. $(z^1)^* = z^* = (z^*)^1$ so our base case holds.

Suppose then that for $n$, $(z^n)^* = (z^*)^n$. We wish to show that $(z^{n+1})^* = (z^*)^{n+1}$. Note that

$$(z^{n+1})^* = (z^nz)^* = (z^n)^*z^*.$$

Can you take it from here?

Answer 3

You should first prove that $$ \overline{ab}=\overline{a}\overline{b}. $$ Next, assuming that $\overline{z^{n}}=\big(\overline{z}\big)^n$ you get $$ \overline{z^{n+1}}=\overline{z^{n}z}=\overline{z^{n}}\overline{z}=\big(\overline{z}\big)^n \overline{z}=\big(\overline{z}\big)^{n+1} $$