e^{πi} + 1 = 0
Which we know from Calculus, Taylor Series and all that jazz. And indeed, every formula of the form:
e^{(2n+1)πi} + 1 = 0; where n ∈ Z
But I digress. Anyway,
e^{πi} + 1 = 0 e^{πi} = -1 e^{2πi} = 1 (Squaring both sides) e^{2πi} = e^{0} (Identity) 2πi = 0 (Stripping away the e's)
Q.E.D.
Answer (highlight):
Well, the answer is, we're playing quick and dirty with the logarithm function. In the crazy world of imaginary numbers the logarithm function isn't a function. While the real part of the function is constant, the imaginary part has an infinite amount of possible values. As you notice the real parts of the equation 2πi = 0 are equal, while the imaginary parts are not. By implicitly taking the logarithm in the last step without regard to imaginary numbers, we violate the rules and we unintentionally throw out a valid solution. Much in the same way as if you have x^{3} + x^{2} + x = 0, and you divide everything by x you throw away the solution x = 0. |