Complex # Huh...
Number Theory
Level
2
Easy but tricky!
Find two distinct numbers such that each number is equivalent to the square of the other number.
[-1 + i√3]/2 and it's conjugate
[-1- i√3]/2 and it's conjugate
w and w^2 ; where w = [-1+i*sqrt(3)]/2 is the complex cube roots of unity.
All of them
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Solution 1:
Let
a is the #, and b is the other #
so,
a = b^2 ; b = a^2
Subtracting them....
a-b = (b+a)(b-a)
(b+a)(b-a) + (b-a) = 0
(b+a+1)(b-a) = 0
(b+a+1) = 0
b = -a-1
b-a = 0
b = a We can't say, b = a because it is already a conjecture on a = b^2
So, since b = -(a+1)
a = b^2
a = [-(a+1)]^2
a = a^2 + 2a + 1
(a^2 + a + 1) = 0
a = [-1±i√3]/2
and b = [-1∓i√3]/2
Generalize: a = [-1 + i√3]/2 and b is just the conjugate of that:
or
a = [-1- i√3]/2 and b is just the conjugate of that:
Solution 2:
Let a and b be the two distinct numbers.
So a = b^2 and b = a^2. Subtracting these equations,
a-b = b^2-a^2,
(b-a)(b+a+1) = 0.
Since b <> a, so b = -1-a. Substituting this expression of b into the second equation,
-1-a = a^2,
a^2+a+1 = 0,
a = w or a = w^2;
b = -1-w = w^2 or b = -1-w^2 = w,
where w = [-1+i*sqrt(3)]/2 is the complex cube roots of unity.