Proving $(\overline{z})^n$ = $\overline{z^n}$ by induction

Question

Prove that $(\overline{z})^n$ = $\overline{z^n}$ for $n \in \mathbb{N}$ by induction.

Answer 1

RESULT: If $z,w\in\Bbb C$ the $\overline{zw}=\overline{z}\cdot\overline{w}$

Problem: $(\overline{z})^n$ = $\overline{z^n}$ for all $n \in \mathbb{N}.$

Proof: (by induction) The result is trivial if $n=1$. Suppose the statement holds for $n=k$. Then $$(\overline{z})^k =\overline{z^k}.$$ Now, $(\overline{z})^k,\overline{z}\in\Bbb R$ and $z^k,z\in\Bbb C$ and hence $$(\overline{z})^{k+1}=(\overline{z})^k\cdot\overline{z}=\overline{z^k}\cdot\overline{z}\overbrace{=}^{\text{use the result}}\overline{z^kz}=\overline{z^{k+1}}.$$ Hence the statement also holds at $n=k+1$. The problem follows by induction.

Answer 2

The base case is trivial $(\overline z=\overline z)$. Assume that $\overline z^k=\overline{z^k}$. Then

$$\overline z^{k+1}=\overline z^k\cdot \overline z=\overline{z^k}\cdot \overline z.$$

Now $z^k$ is some complex number $a+bi$ and $z$ is another complex number $c+di$.

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

and

$$\overline{z^{k+1}}=\overline{ac-bd+(ad+cb)i}$$