How does $x+iy = (s+it)^n \implies x-iy = (s-it)^n,$ where $x, y, s, t\in\Bbb R?$

Question

How does $x+iy = (s+it)^n \implies x-iy = (s-it)^n,$ where $x, y, s, t$ are real numbers? (This is only a fragment of the problem I'm solving, but the only part I'm stuck on is this one.) Why it still remains equal even after conjugating?

Thanks!

Answer 1

$$x-iy=\overline {x+iy}=\overline {(s+it)^n}=(\overline{s+it})^n=(s-it)^n$$

Answer 2

For $z\in\mathbb{C}$ and $n\in\mathbb{N}, \overline{z^n}=\overline{z}^n$. It can be proved by writing $z=re^{i\theta}$.

Answer 3

Use the binomial theorem: $$ (s+it)^n = \sum_{k=0}^n \binom{n}{k} s^{n-k} t^k i^{k} , $$ and taking the complex conjugate, $ \overline{i^k} = (-i)^{k} $ (you can check this for all four distinct powers of $i$ if you don't believe it, or use induction), and the result now follows since $$ \sum_{k=0}^n \binom{n}{k} s^{n-k} t^k (-i)^{k} = (s-it)^k . $$