Complex Numbers (1 Viewer)

[Vision]

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]

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.

 

[esaitchkay]

EDIT: nvm already answered! haha too quick

 

[Vision]

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]

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?

 

[Carrotsticks]

Sorry, how did you get 1+w+w^2=0 just from w^3=1?

w^3=1

w^3-1=0

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

Since w is the non-real root, then w=/=1.

Hence, 1+w+w^2=0.

 

[bottleofyarn]

Sorry, how did you get 1+w+w^2=0 just from w^3=1?

You can factorise using . Since w =/= 1 then

 

[ALL_CAPS]

w^3=1

w^3-1=0

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

Since w is the non-real root, then w=/=1.

Hence, 1+w+w^2=0.

Ahh, thanks, didnt expect to use factorisation :0, guess i need to get used to these question types

 

[hit patel]

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]

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

 

[Kurosaki]

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

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?

 

[Chicken Burger]

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!

 

[Kurosaki]

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!

4ix=0 because if is real then there can be no imaginary component.

 

[Chicken Burger]

4ix=0 because if is real then there can be no imaginary component.

oh yeah yeah true, ok thanks friend