I had these questions in my 4 unit maths test yesterday, the red question is the one that I got wrong however I cannot see a problem with my working:

 

a) Find the non real roots of x³ + 1 = 0

 

[math]x = \frac{1}{2} \pm \frac{\sqrt{3}}{2} i[/math]

 

b) If h is one of those roots prove that h² = h - 1

 

[math]LHS = h^2 = \frac{1}{4} \pm \frac{\sqrt{3}}{2}i - \frac{3}{4} = \frac{-1}{2} \pm \frac{\sqrt{3}}{2} i = h - 1 = RHS[/math]

 

c) Hence, find the value of (1-h)^6

 

[math](1-h)^6 = 1 - 6h + 15h^2 - 20h^3 + 15h^4 - 6h^5 + h^6[/math]

 

Considering the roots of unity:

 

[math]= 1 - 6h + 15h^2 - 20 + 15h - 6h^2 + 1[/math]

 

[math]= -18 + 9h + 9h^2[/math]

 

[math]= -27[/math]

 

She got an answer of 1. Why is mine wrong?

Because it is: i presume you're subbing h^2=h-1 in at some point. Why not do it at the start instead of expanding a sextic unnecessarily? (you're algebra is just wrong)

I had these questions in my 4 unit maths test yesterday' date=' the red question is the one that I got wrong however I cannot see a problem with my working:

 

[b']a) Find the non real roots of x³ + 1 = 0[/b]

 

[math]x = \frac{1}{2} \pm \frac{\sqrt{3}}{2} i[/math]

 

b) If h is one of those roots prove that h² = h - 1

 

[math]LHS = h^2 = \frac{1}{4} \pm \frac{\sqrt{3}}{2}i - \frac{3}{4} = \frac{-1}{2} \pm \frac{\sqrt{3}}{2} i = h - 1 = RHS[/math]

 

c) Hence, find the value of (1-h)^6

 

[math](1-h)^6 = 1 - 6h + 15h^2 - 20h^3 + 15h^4 - 6h^5 + h^6[/math]

 

Considering the roots of unity:

 

[math]= 1 - 6h + 15h^2 - 20 + 15h - 6h^2 + 1[/math]

 

[math]= -18 + 9h + 9h^2[/math]

 

[math]= -27[/math]

 

She got an answer of 1. Why is mine wrong?

Yeah since you've proven B you can make

[math](1-h)^{6}[/math]

into

[math](-1)^{6}(h^{2})^{6} = h^{12} = (h^{3})^{4} = (-1)^{4} = 1[/math]

Because it is: i presume you're subbing h^2=h-1 in at some point. Why not do it at the start instead of expanding a sextic unnecessarily? (you're algebra is just wrong)

 

Where is my algebra wrong? The expansion looks correct, then I subtract 3 from any indicies which are greater than 2 because the roots repeat, then I subtract 9 of each root since the roots sum to zero. That is the technique I have been taught to use with roots of unity.

 

cosine, yeah that was the solution she gave me.

 

I would appreciate to know exactly where the error is because even my Maths teacher couldn't work out why my answer was wrong.

The roots don't repeat with period 3.

 

They aren't cube roots of unity.

 

They are cube roots of -1.

 

h^4=h^3h=-h.

 

They satisfy x^3+1=0, not x^3-1=0.

The roots don't repeat with period 3.

 

They aren't cube roots of unity.

 

They are cube roots of -1.

 

h^4=h^3h=-h.

 

They satisfy x^3+1=0' date=' not x^3-1=0.[/quote']

 

Agh! Why didn't I see that? Thanks for pinpointing the error Matt.