Where does complex exponential come from?

The complex exponential function is defined as : $$e^{ix} = \cos x + i\sin x$$ It shares most of its properties with real exponential and it allows a lot of trigonometric calculations such as de Moivre's formula : $$(\cos x+i\sin x)^n = \cos{nx}+i\sin{nx}$$

But where does this definition come from and why does it work ?

Answer

Another way to look at it is to view the exponential and trigonometric functions as defined by a power series:

$$\exp(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$$

$$\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$$

$$\cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots$$

This has the advantage that the $x$ can be anything, as long as we know how to multiply two of them, add two of them together, and divide them by a real number. In particular, it makes sense for both real and complex numbers.

Now you can put $ix$ into the definitions in place of $x$, and compute:

$$\begin{align}
\exp(ix)
& = 1 + ix + \frac{(ix)^2}{2!} + \frac{(ix)^3}{3!} + \cdots \\
& = \left( 1 - \frac{x^2}{2!} + \cdots\right) + i \left( x - \frac{x^3}{3!} + \cdots\right) \\
& = \cos x + i \sin x
\end{align}$$

so the formula you quoted is seen to be a theorem rather than a definition. If we now assume that the familiar law

$$\exp(a+b) = \exp(a) \exp(b)$$

holds for arbitrary $a$ and $b$ (it does, and you can prove it from the power series definition) then we now have a way to compute the exponential of any complex number:

$$\exp(x+iy) = \exp(x) (\cos y + i\sin y)$$

where $x$ and $y$ are real.