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jkasdhk
If α and β are complex cube roots of unity then (α^4 β^4)+1/αβ=?
the product of the roots αβ=1 and (α^3 β^3)=1
We see sum of roots and product of roots?
\[\alpha * \beta \] isnt equal to 1....cube root remember?
the answer given is 0. i dont know how?
infact the answer is zero
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....
complex cube roots means α=β?
do u agree the roots \[\alpha =\Omega and \beta = \Omega ^{2}\]? or vice versa
no, not necessarily all the time. for unity yes.
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
\[ \alpha=e^{\frac{2 i \pi }{3}}; \quad \alpha^3 =1\\ \beta =1; \quad \beta^3 =1\\ \alpha \beta \ne 1 \]
@amrit110: can u please tell me the steps to solve this ques?
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.
note that the complex cube roots of unity are the zeroes of the eqn x^2 + x +1 so (alpha)(beta) = 1 try now..
There are 3 complex roots of unity: \[ \left\{1,e^{\frac{2 i \pi }{3}},e^{-\frac{2 i \pi }{3}}\right\} \]
A complex root of unity is a root of \[ x^3 =1 \]
@eliassaab i didn't get how come αβ not 1
@amrit110: What will be the answer of (α^4 β^4) ?
If both \( \alpha \ne 1 \) and \( \beta \ne 1 \) then \(\alpha \beta =1\)
In the above case the ratio is 2.
it should be 1 jkasdhk
yes for complex nos .. for real nos If both α≠1 and β≠1 then αβ not 1
but 2 is not the answer
is the question (α^4 β^4) -1/αβ or still different
If we consider complex roots only and \(\alpha \ne \beta\) then \[ \alpha = e^{\frac{2 i \pi }{3}}\\ \beta = e^{-\frac{2 i \pi }{3}}\\ \alpha \beta= e^0 =1 \]
If the question is: Let \( \alpha, \beta\) be the 2 complex roots (not real), then the ratio in question is 2
@amrit110: How is (α^4 β^4)=1
\[ \alpha^4 \beta^4 = \alpha^3 \beta^3 \alpha \beta = (1)(1) =\alpha \beta =1 \] If \( \alpha\ne \beta \)
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
This product is \[ (-1)^3 \frac {-1}{1}=1 \] See http://en.wikipedia.org/wiki/Vieta%27s_formulas