Proof (-3)squared is -3!!!! Not 9

Start of proof:

Euler's formula that I rediscovered when I was in the 10th grade when I learned



e^(pi*i) = -1



Then assumed rightly when I was in pre-calc that the pi is similar to the radians on the sine and cosine functions



Derived from my TI83 graphing calculator in the simple days before you could look anything up online



That e^(x*i) = cosine x + i*sine x



Sure enough I was right

http://ift.tt/1vRSHIG



But I now found a fuzzy area of this formula



If e to the pi i = -1



Then e to the (pi i)/3



= -1 since that's the cube root of -1



And -1^3 = -1



So that would be e to the x*i

In the case where x = pi/3



e^((Pi/3)*i) = (e^(pi*i)*(1/3))

= (-1)^(1/3)

=-1



Pi/3 radians is angle 60 degrees in deg mode



That's



Cosine 60degrees + i Sine 60degreee



We established this is the cube root of -1 which is -1



-1

=

1/2. + i times sqrt(3)/2



Hence



1 + sqrt (-3) = -2



Sqrt (-3) = -3



Square both sides



-3 = (-3)^2



What am I missing?

Proof (-3)squared is -3!!!! Not 9