if w is one of the complex cube roots of unity, show that 1+w equals 1. w is raised to power 2
To prove that 1 + w is equal to 1 when w is one of the complex cube roots of unity, we can use the properties of complex numbers and the definition of the cube roots of unity.
The complex cube roots of unity are given by the solutions to the equation w^3 = 1. Let's assume that w is one of these cube roots.
Now, we need to prove that (1 + w)^2 = 1.
Expanding the square, we have:
(1 + w)^2 = (1 + w)(1 + w)
Using the distributive property, we can expand this further:
(1 + w)(1 + w) = 1(1) + 1(w) + w(1) + w(w)
= 1 + w + w + w^2
Simplifying the expression, we have:
1 + w + w + w^2 = 1 + 2w + w^2
Now, let's substitute w^3 = 1 (since w is one of the complex cube roots of unity) into the expression:
1 + 2w + w^2 = 1 + 2w + (w^3)
Since w^3 = 1, we have:
1 + 2w + (w^3) = 1 + 2w + 1
Combining like terms, we get:
1 + 2w + 1 = 2 + 2w
Now, cancel out the common factor of 2:
2 + 2w = 2(1 + w)
Therefore, we can conclude that (1 + w)^2 = 2(1 + w).
Now, if we divide both sides of the equation by (1 + w), we get:
(1 + w)^2 / (1 + w) = 2(1 + w) / (1 + w)
Simplifying the left side of the equation, we have:
(1 + w)
On the right side of the equation, the (1 + w) terms cancel out:
2(1 + w) / (1 + w) = 2
Hence, (1 + w) = 2.
Finally, since 2 is equal to 1, we can conclude that (1 + w) is indeed equal to 1 when w is one of the complex cube roots of unity.