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Find the flaw with this proof: sqrt(-1)<0

Question

Object: to prove that i < 0 ( that is, sqrt(-1) < 0 )

( .5 + sqrt(3/4)*i )^3 = (-1)^3

which means that .5 + sqrt(3/4)*i = -1

So then 1 + sqrt(3)*i = -2

sqrt(3)*i = -1

i = -1/sqrt(3)

Therefore i is a negative number. QED.

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Object: to prove that i < 0 ( that is, sqrt(-1) < 0 )

( .5 + sqrt(3/4)*i )^3 = (-1)^3

which means that .5 + sqrt(3/4)*i = -1

So then 1 + sqrt(3)*i = -2

sqrt(3)*i = -1

i = -1/sqrt(3)

Therefore i is a negative number. QED.

when talking about complex number, there are 2 things, magnitude and arguement 0.5+sqrt(3/4)*i 's magnitude is equal to one and the arguement results in the further -ve sign. as aruement is pi/3 and arguement^3=3*arguement(a property whose name i have forgetton i.e. arguement^n=n*arguement)=pi. and 1*pi =-1 in complex.

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Obvious error in the proof is the part "which means that .5 + sqrt(3/4)*i = -1".. Generally when x^3=y^3, you can never directly say x is always equal to y.. It only leads to the conclusion that (x^3-y^3)=0, which on factorisation leads to (x-y)*(x^2+xy+y^2)=0.. So any further solution after that particular statement is not accepted in algebra..

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