TITLE: Given $e^{i\theta}=x+iy$, how do you prove that $e^{-i\theta}=x-iy$ without using Euler's formula?
QUESTION [6 upvotes]: Suppose we use the following definition of complex exponentiation:
$$
e^z=\lim_{n \to\infty}\left(1+\frac{z}{n}\right)^n \, .
$$
For a given real value of $\theta$, let $e^{i\theta}=x+iy$. I want to show that $e^{-i\theta}=x-iy$ without using Euler's formula.
I think that this is true because the algebraic properties of $i$ and $-i$ are identical. This means, roughly, that if we have a valid equation involving $i$, and replace every occurence of $i$ with $-i$, then we still get a valid equation. Here is my attempt to turn it into a rigorous argument. Is it correct?

Since the mapping $a+bi\mapsto a-bi$ is a field automorphism of $\Bbb{C}$, we know that for all $n\in\Bbb{N}$,
$$
\left(\overline{1+\frac{i\theta}{n}}\right)^n=\overline{\left(1+\frac{i\theta}{n}\right)^n} \, .
$$
Hence,
\begin{align}
e^{-i\theta} &= \lim_{n \to \infty}\left(1-\frac{i\theta}{n}\right)^n \\[5pt]
&= \lim_{n\to\infty}\left(\overline{1+\frac{i\theta}{n}}\right)^n \\[5pt]
&= \lim_{n\to\infty}\overline{\left(1+\frac{i\theta}{n}\right)^n}  \\[5pt]
&= \overline{\lim_{n\to\infty}\left(1+\frac{i\theta}{n}\right)^n} \tag{*}\label{*}\\[5pt]
&= \overline{e^{i\theta}} \\[5pt]
&= x-iy \, .
\end{align}
The line $\eqref{*}$ is justified because the function $a+bi\mapsto a-bi$ is continuous.

REPLY [9 votes]: It seems fine, maybe we can also proceed as follows, since
$$e^{i\theta}\cdot e^{-i\theta}=e^0=1$$
we have
$$(x+iy)(a+ib)=ax-by+i(ay+bx)=1$$
and

$ay+bx=0 \implies a=-b\frac x y$
$ax-by=-b\frac {x^2}y-by=-\frac b y=1 \implies b=-y \quad a=x$