Prove e^(ix) = cos (x) + isin(x)

We first write each side of the equation using the maclaurin series for each function.

e^ix = 1 + ix + (ix)²/2! + (ix)³/3! + (ix)⁴/4! + ......

e^ix = 1 + ix - x²/2! - ix³/3! + x⁴/4! + .....

cos(x) + isin(x) = (1 - x²/2! + x⁴/4! - x⁶/6! +....) + i(x - x³/3! + x⁵/5! - x⁷/7! + ......)

writing the above equation in increasing powers of x:

cos(x) + isin(x) = 1 + ix - x²/2! - ix³/3! + x⁴/4! + ....

As seen the maclaurin series for each side of the equation are the same hence e^ix = cos(x) + isin(x)