Why does e^ix = cos(x) + isin(x)

If you look at the Taylor series expansion of e^x: e^x = 1 + x + x²/(2!) + x³/(3!) + x⁴/(4!)...

If you then make this e^ix, you get

1+ix - x²/(2!) - ix³/(3!) + x⁴/(4!)...

If we split this into real and imaginary, we see the real part is

1 - x²/(2!) + x⁴/(4!) - x⁶/(6!)...

The series expansion of cos(x) is

1 - x²/(2!) + x⁴/(4!) - x⁶/(6!)...

Therefore, the real part of e^ix is cos(x)

If we look at only the imaginary part of e^ix, we get

i(x - x³/(3!) + x⁵/(5!) - x⁷ / (7!)

If we look at the series expansion of sin(x) we get

(x - x³/(3!) + x⁵/(5!) - x⁷ / (7!)

Therefore the imaginary part of e^ix = sin(x)

Putting this together, we get e^ix = Re(e^ix) + Im(e^ix)

e^ix = cos(x) + isin(x)