CLarification need please for complex numbers (1 Viewer)

[HeroicPandas]

for this question

If α is a complex root of z⁷-1=0, show that:

a) α³ is also a complex root and hence state all the complex roots of z⁷-1=0 in terms of α. (answer = α, α², α³,α⁴,α⁵,α⁶)

b) The quadratic equation whose roots are α+α²+α⁴ and α³+α⁵+α⁶ is z²+z+2=0


for part a)
can i write: if α is a complex number α must satisfy z^7-1=0

hence, therefore

α^7-1=0
factorise
(α-1)(1+α+α^2+...+α^6)=0

if α is a root of z^7-1=0, therefore a^2, a^3, a^4, a^5, a^6 are also roots. AS REQUIRED

α is a complex, therefore a cannot eual to 1

can i write this in the hsc?

 

[math man]

no that doesnt prove those are the other roots, it only proves their sum is 0.

 

[D94]

What you have shown isn't greatly relevant to the first part.

You need to find the roots not by factorising, but by solving z⁷-1=0. Then you will see that your first solution will be z = 1, then your next solution is z = a, and then z = a² and so on until z = a⁶, which are your required roots.

 

[HeroicPandas]

dam then i gotta to that way...i was trying to find out a different method

 

[D94]

dam then i gotta to that way...i was trying to find out a different method

The generally preferred method is rather quick though. With practice, and of course understanding what you're doing, it will become second nature to quickly solve the roots of unity.

 

[HeroicPandas]

The generally preferred method is rather quick though. With practice, and of course understanding what you're doing, it will become second nature to quickly solve the roots of unity.

aight cool thanks

 

[asianese]

Do we need to go into details like this?

 

[Carrotsticks]

Looks good.

But I would probably specify:



You also left out the trivial solution. I know the Q just wants complex solutions, but you omitted it without explanation.

 

[karnbmx]

Looks good.

But I would probably specify:



You also left out the trivial solution. I know the Q just wants complex solutions, but you omitted it without explanation.

How can have multiple values?

I think what you meant to say was 1 <= n < 7 =)

 

[Carrotsticks]

How can have multiple values?

I think what you meant to say was 1 <= n < 7 =)

Because there are 7 solutions, by the Fundamental Theorem of Algebra, so there are 7 solutions.

ie: 7 values of theta.

 

[RealiseNothing]

How can have multiple values?

I think what you meant to say was 1 <= n < 7 =)

Subscript 'i', not imaginary unit 'i'.