## jkasdhk 3 years ago If α and β are complex cube roots of unity then (α^4 β^4)+1/αβ=?

1. matricked

is it 2

2. jkasdhk

How?

3. matricked

the product of the roots αβ=1 and (α^3 β^3)=1

4. jkasdhk

How is (α^3 β^3)=1?

5. jkasdhk

We see sum of roots and product of roots?

6. amrit110

$\alpha * \beta$ isnt equal to 1....cube root remember?

7. jkasdhk

the answer given is 0. i dont know how?

8. mayank_mak

9. jkasdhk

How????

10. amrit110

well its complex cube root, $\alpha = \beta$ = $\sqrt{-1}$. so that would make cube of it equal to 1. but the other term -1. 1-1=0....

11. jkasdhk

complex cube roots means α=β?

12. mayank_mak

do u agree the roots $\alpha =\Omega and \beta = \Omega ^{2}$? or vice versa

13. amrit110

no, not necessarily all the time. for unity yes.

14. matricked

if α and β are the complex cube roots of 1 then α and β must satisfy the eq. x^2 +x +1=0 and x^3=1 thus α + β=-1 and αβ=1

15. eliassaab

$\alpha=e^{\frac{2 i \pi }{3}}; \quad \alpha^3 =1\\ \beta =1; \quad \beta^3 =1\\ \alpha \beta \ne 1$

16. jkasdhk

@amrit110: can u please tell me the steps to solve this ques?

17. mayank_mak

$\Omega ^{3} = 1$

18. amrit110

ok. see complex cube roots of unity so $x^{3}=1$. so when u solve for x it is obviously the root of (-1). now this means that for this case we can safely assume alpha n beta to be equal to x. although this is not true for all cubic equations, sometimes the roots can be complex conjugate too. but it doesnt matter.

19. shubhamsrg

note that the complex cube roots of unity are the zeroes of the eqn x^2 + x +1 so (alpha)(beta) = 1 try now..

20. eliassaab

There are 3 complex roots of unity: $\left\{1,e^{\frac{2 i \pi }{3}},e^{-\frac{2 i \pi }{3}}\right\}$

21. eliassaab

A complex root of unity is a root of $x^3 =1$

22. matricked

@eliassaab i didn't get how come αβ not 1

23. jkasdhk

@amrit110: What will be the answer of (α^4 β^4) ?

24. eliassaab

If both $$\alpha \ne 1$$ and $$\beta \ne 1$$ then $$\alpha \beta =1$$

25. eliassaab

In the above case the ratio is 2.

26. amrit110

it should be 1 jkasdhk

27. matricked

yes for complex nos .. for real nos If both α≠1 and β≠1 then αβ not 1

28. jkasdhk

and 1/αβ would be?

29. matricked

1/αβ would be 1

30. jkasdhk

1+1 will be 2? right?

31. jkasdhk

but 2 is not the answer

32. matricked

is the question (α^4 β^4) -1/αβ or still different

33. eliassaab

If we consider complex roots only and $$\alpha \ne \beta$$ then $\alpha = e^{\frac{2 i \pi }{3}}\\ \beta = e^{-\frac{2 i \pi }{3}}\\ \alpha \beta= e^0 =1$

34. jkasdhk

its (α^4 β^4) +1/αβ

35. eliassaab

If the question is: Let $$\alpha, \beta$$ be the 2 complex roots (not real), then the ratio in question is 2

36. jkasdhk

@amrit110: How is (α^4 β^4)=1

37. eliassaab

$\alpha^4 \beta^4 = \alpha^3 \beta^3 \alpha \beta = (1)(1) =\alpha \beta =1$ If $$\alpha\ne \beta$$

38. eliassaab

Let $$1, \alpha, \beta$$ be the three roots of unity,then $\frac { \alpha ^4 \beta^4 +1}{\alpha \beta }= 2$ It is easy to see now $(1) \alpha \beta =1$ Since the last product is the product of the three roots of the polynomial $x^3 -1=0 39. eliassaab This product is \[ (-1)^3 \frac {-1}{1}=1$ See http://en.wikipedia.org/wiki/Vieta%27s_formulas

40. eliassaab

@jkasdhk