How can an imaginary equation give a real answer?

Question

I came across this equation,
$$ e^{ix} = \cos(x) + i\sin(x) $$ This is the simplified version, the real one is more complex but this part is the one I have a question about.
The right side clearly has a $i$ term in it, and is thus imaginary, but the left, $e^{x}$ is entirely real. How is this feasible? I couldn't find many good sources on the derivation of this equation, so maybe I am missing something?

NOTE: I came across the equation from a physics source, so I am not extremely adept with the mathematics behind this.

Answer 1

$e^y$ is a function whose range (output) are positive real numbers when the real numbers is its domain. However when we say $e^{ix}$ we are using the function $e^y$ on the complex number $ix$, thus we suddenly have a bit of a different function, acting on complex numbers, and thus we do not necessarily need to have a real number as output.

Answer 2

This identity can be derived using the Taylor Series expansions for $e^{ix}$, $\sin(x)$, and $\cos(x)$:

$$\cos(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots$$ $$i\sin(x)=i(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots)$$ $$e^{ix}=1+ix+\frac{(ix)^2}{2!}+\frac{(ix)^3}{3!}+\cdots=1+ix-\frac{x^2}{2!}+\frac{ix^3}{3!}+\frac{x^4}{4!}-\cdots$$

Adding the series for $\cos(x)$ and $i\sin(x)$ together, we obtain the the series for $e^{ix}$.