Prove that $\overline{e^{ix}}=e^{-ix}$ I am trying to construct properly the imaginary exponential as an extension of the real exponential. For this, I need to show that  $\overline{e^{ix}}=e^{-ix}$.
I know that there is a very simple proof using an infinite sum of powers. But I am really looking for something purely algebraic using mainly if not only the properties of the real exponential extended to complex numbers.
One can easily show that $1=e^{ix}\cdot e^{-ix}$ and therefore that if $z=e^{ix}$, then $e^{-ix}=\lambda\cdot \overline z$ with $\lambda\in\mathbb R$, but how can we show that $\lambda=1$ ?
Once again, I am really looking for a proof that doesn't use any infinite sum.

Just use Euler's formula:
$$\overline{e^{ix}}=\overline{\cos x+i\sin x}=\cos x-i\sin x=\cos(-x)+i\sin(-x)=e^{-ix}.$$
