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Complex Numbers (1 Viewer)

Vision

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Hey guys,

Been stuck on this question for some time now, but I have a feeling i'm over-thinking it.

If w is a complex cube root of unity, show that:

(1+w-w^2)^3 - (1-w+w^2)^3 = 0

Thanks :)
 

Carrotsticks

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If w is a complex cube root of unit, then 1+w+w^2=0.

So we have 1-w+w^2 = -2w, and 1+w-w^2=-2w^2.

Hence, the LHS is (-2w)^3 - (-2w^2)^3 = -8w^3 + 8w^6 = -8+8 (since w^3=1) and we thus have 0.
 

Vision

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If w is a complex cube root of unit, then 1+w+w^2=0.

So we have 1-w+w^2 = -2w, and 1+w-w^2=-2w^2.

Hence, the LHS is (-2w)^3 - (-2w^2)^3 = -8w^3 + 8w^6 = -8+8 (since w^3=1) and we thus have 0.
Oh right I see, thank you!
 

ALL_CAPS

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If w is a complex cube root of unit, then 1+w+w^2=0.

So we have 1-w+w^2 = -2w, and 1+w-w^2=-2w^2.

Hence, the LHS is (-2w)^3 - (-2w^2)^3 = -8w^3 + 8w^6 = -8+8 (since w^3=1) and we thus have 0.
Sorry, how did you get 1+w+w^2=0 just from w^3=1?
 

hit patel

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or a more complicated way:
(w^2)^3=(w^3)^2=1
(w)^3 =1
Therefore w,1, w^2 are all roots of the equation w^3-1=0
Therefore
Therefore Σ A: w+W^3+1=-b/a = 0.
But yeh the factorisation is much easier unless asked for the use of polynomial entities.
 

Chicken Burger

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hi guys i need help with this one:

z is any number, such that Im(z) = 2 and z^2 is real. Find z

thanks
 

Chicken Burger

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Using z=x+iy, since Im(z)=2, then z=x+2i.
Squaring, we get . Since z^2 is real, 4ix=0, therefore x=0.
therefore z=2i?
thanks kuro that is correct, im kinda confused in "since z^2 is real, 4ix=0" isnt "4ix" the imaginary because of the i ? thank you!
 

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