## jkasdhk Group Title If α and β are complex cube roots of unity then (α^4 β^4)+1/αβ=? 2 years ago 2 years ago

1. matricked Group Title

is it 2

2. jkasdhk Group Title

How?

3. matricked Group Title

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

4. jkasdhk Group Title

How is (α^3 β^3)=1?

5. jkasdhk Group Title

We see sum of roots and product of roots?

6. amrit110 Group Title

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

7. jkasdhk Group Title

the answer given is 0. i dont know how?

8. mayank_mak Group Title

9. jkasdhk Group Title

How????

10. amrit110 Group Title

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 Group Title

complex cube roots means α=β?

12. mayank_mak Group Title

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

13. amrit110 Group Title

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

14. matricked Group Title

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 Group Title

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

16. jkasdhk Group Title

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

17. mayank_mak Group Title

$\Omega ^{3} = 1$

18. amrit110 Group Title

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 Group Title

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 Group Title

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

21. eliassaab Group Title

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

22. matricked Group Title

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

23. jkasdhk Group Title

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

24. eliassaab Group Title

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

25. eliassaab Group Title

In the above case the ratio is 2.

26. amrit110 Group Title

it should be 1 jkasdhk

27. matricked Group Title

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

28. jkasdhk Group Title

and 1/αβ would be?

29. matricked Group Title

1/αβ would be 1

30. jkasdhk Group Title

1+1 will be 2? right?

31. jkasdhk Group Title

but 2 is not the answer

32. matricked Group Title

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

33. eliassaab Group Title

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 Group Title

its (α^4 β^4) +1/αβ

35. eliassaab Group Title

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

36. jkasdhk Group Title

@amrit110: How is (α^4 β^4)=1

37. eliassaab Group Title

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

38. eliassaab Group Title

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 Group Title This product is \[ (-1)^3 \frac {-1}{1}=1$ See http://en.wikipedia.org/wiki/Vieta%27s_formulas

40. eliassaab Group Title

@jkasdhk