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.
 A: Just use Euler's formula:
$$\overline{e^{ix}}=\overline{\cos x+i\sin x}=\cos x-i\sin x=\cos(-x)+i\sin(-x)=e^{-ix}.$$
A: How should you define $e^z$ for complex numbers so that it retains the good properties of the real function $e^x$?  You take a definition of $e^x$ for real $x$ and modify it to apply to complex values.  Then prove all the properties you want.
For each definition of the real case, you get a corresponding definition for the complex case.
One definition is the Taylor series.  So for complex $z$, define
$$
F(z) = \sum_{n=0}^\infty \frac{z^n}{n!},\quad z\in \mathbb C
$$
Another  definition is the differential equation.  So for complex $z$, define $F(z)$ to be the unique solution $F : \mathbb C \to \mathbb C$ for the differential equation
$$
F'(z)=F(z),\quad F(0)=1
$$
If you like another definition, try to use it...

Summary: for your original question, the answer will depend on which definition you used.
A: Write $e^{ix} = a + bi$ where $a, b \in \mathbb{R}$. Clearly $\overline{e^{ix}} = a-bi$. Notice now the following:
$$e^{-ix} = \frac{1}{e^{ix}} = \frac{1}{a+bi} = \frac{a-bi}{a^2 + b^2} = \frac{\overline{e^{ix}}}{a^2 + b^2}$$
Hence $(a^2 +b^2)e^{-ix} = \overline{e^{ix}}$. Since $a^2 + b^2 = 1$, the result follows.
Edit: the fact that $|e^{ix}| = a^2 + b^2 = 1$ can be proven without using infinite sums or Euler's formula, as is done here.
A: Let $P(f) = f(x + iy) - f(x-iy)$ and $H(f) = f(x+iy) + f(x-iy)$. These transformations are linear, i.e. $P(\alpha f + \beta g) = \alpha P(f) + \beta P(g)$, and $H(\alpha f + \beta g) = \alpha H(f) + \beta H(g)$.
Now, for $x$ real, we have:
$$
\overline{e^x} = e^{\overline{x}}.
$$
Applying $P$ on the LHS yields
$$
P[\overline{e^x}] = \overline{P[e^x]} = \overline{e^{x+iy}} - \overline{e^{x-iy}},
$$
and similarly on the RHS
$$
P[e^{\overline{x}}] = e^{\overline{x+iy}} - e^{\overline{x-iy}} = e^{x-iy} - e^{x+iy}.
$$
Thus:
$$
\overline{e^{x+iy}} - \overline{e^{x-iy}} = e^{x-iy} - e^{x+iy} \implies \overline{e^{x+iy}} + e^{x+iy} = \overline{e^{x-iy}} + e^{x-iy} \implies \Re{e^{x+iy}} = \Re{e^{x-iy}}
$$
Applying $H$ on the LHS we get
$$
H[\overline{e^x}] = \overline{H[e^x]} = \overline{e^{x+iy}} + \overline{e^{x-iy}},
$$
and similarly on the RHS
$$
H[e^{\overline{x}}] = e^{\overline{x+iy}} + e^{\overline{x-iy}} = e^{x-iy} + e^{x+iy},
$$
so
$$
\overline{e^{x+iy}} + \overline{e^{x-iy}} = e^{x-iy} + e^{x+iy}\implies e^{x+iy} - \overline{e^{x+iy}} = \overline{e^{x-iy}} - e^{x-iy}  \implies \Im{e^{x-iy}} = -\Im{e^{x+iy}}.
$$
Therefore for $z = x+iy$, $e^z$ is a conjugate of $e^{\overline{z}}$, i.e. $\overline{e^z} = e^{\overline{z}}$. The result $\overline{e^{ix}}=e^{-ix}$ follows trivially.
