Complex Cube Root Of Unity (1 Viewer)

[ADrew]

Hey all , can't figure out these two questions, was wondering if some kind person (spiral im looking at you ) may help :

1) If 1, w, w^2 are the three cube roots of unity, prove that:
(a+b+c)(a+bw+cw^2)(a+bw^2+cw)=a^3+b^3+c^3-3abc

And I know the following isnt a root of unity type question, but anyway:

2) Find x, where 0<=x<=2pi, if (3+2i sinx)/(1-2i sinx) is purely imaginary.

Thankyou for any help in advanced!!

 

[IamBread]

Haven't done question 1, but here's 2.















Will now attempt the next one.

 

[SpiralFlex]

What's up duck?! (bugs bunny!) *RAWR!*

Want the solution or a hint?

 

[ADrew]

Thanks IamBread! SpiralFlex, just post whatever you feel.

 

[IamBread]

Spiral do you know how to do the plusminus symbol?

 

[SpiralFlex]

Thanks IamBread! SpiralFlex, just post whatever you feel.

Note:

 

[ADrew]

Thanks so much! Appreciate your response
BTW rep to both of you

 

[IamBread]

Haha, no I had got it, was just looking for a better way, did not want to latex it lol

 

[SpiralFlex]

I'm going to sleep now, hopefully i'll get some study done once I wake up and not play games.

Be back at 6.

To infinity and beyond!

Swoosh!

 

[RealiseNothing]

Alternatively:

The 3 roots are equally spaced around the unit circle (any complex circle for that matter).

So the first root = 1, second root (w) = cis120 , and third root (w^2) = cis240

Working out each root and putting it into a+bi form:

cis120 = -1/2 + (rt3)i/2 = w

cis240 = -1/2 - (rt3)i/2 = w^2

Substitute these values into the original equation. Solve the last two brackets first as they are more complicated.

We can actually do a little tampering with the signs of the roots, so that w = w^2. The imaginary part of w is actually the co-efficient of (b-c) and the imaginary part of w^2 is the co-efficient of (c-b). So by taking out a negative from (c-b), we get the negative of the imaginary part of w^2 as the co-efficient of (b-c). But the negative of the imaginary part of w^2 is w. So now w=w^2.

This leaves you with the two last brackets as just a difference of two squares. When you solve this it simplifies to a^2 + b^2 + c^2 - ab - ac - bc. All you need to do now is multiply this with the very first bracket (a+b+c) which gives the answer of:

a^3 + b^3 + c^3 - 3abc.

QED.

This method cancels out all the 'i' from the equation when you use the difference of two squares as they just become -1, leaving you with only a, b, and c to work with. More of an algebraic approach.

 

[ADrew]

Just one more thing, but why does w^4=w ? I realise this may be obvious, but I cant see it?
Also, thanks for the response RealiseNothing

 

[SpiralFlex]

Just one more thing, but why does w^4=w ? I realise this may be obvious, but I cant see it?

 

[kingkong123]

Just one more thing, but why does w^4=w ? I realise this may be obvious, but I cant see it?

lol w^4 = (w^3) x w
but w^3 = 1
therefore w^4 = 1 x w
= w

 

[RealiseNothing]

Just one more thing, but why does w^4=w ? I realise this may be obvious, but I cant see it?

The roots of a complex number can be graphed around a circle on the argand diagram right? And you know how they are equally spaced around it?

Well when you get to the last root (in this case w^2), you go back to the beginning where your first root is, so w^3=1.

Then you repeat the cycle, so w^4=w, w^5=w^2, and so on.

 

[SpiralFlex]

I'm going for my nap nap now.

 

[kingkong123]

beaten to it :/

 

[ADrew]

Lolz, basic algebra escaped my mind for a moment there (facepalm) Anyway thanks again

 

[Nooblet94]

Spiral do you know how to do the plusminus symbol?

\pm, also the version where minus is on top is \mp

 

[Carrotsticks]

Spiral do you know how to do the plusminus symbol?

\pm

 

[IamBread]

\pm

\pm, also the version where minus is on top is \mp

thanks!