- jkasdhk

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

- katieb

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- anonymous

is it 2

- jkasdhk

How?

- anonymous

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

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## More answers

- jkasdhk

How is (α^3 β^3)=1?

- jkasdhk

We see sum of roots and product of roots?

- anonymous

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

- jkasdhk

the answer given is 0. i dont know how?

- anonymous

infact the answer is zero

- jkasdhk

How????

- anonymous

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....

- jkasdhk

complex cube roots means α=β?

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- jkasdhk

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

- anonymous

\[\Omega ^{3} = 1\]

- anonymous

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.

- 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..

- anonymous

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

- anonymous

A complex root of unity is a root of
\[
x^3 =1
\]

- anonymous

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

- jkasdhk

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

- anonymous

If both \( \alpha \ne 1 \) and \( \beta \ne 1 \) then \(\alpha \beta =1\)

- anonymous

In the above case the ratio is 2.

- anonymous

it should be 1 jkasdhk

- anonymous

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

- jkasdhk

and 1/αβ would be?

- anonymous

1/αβ would be 1

- jkasdhk

1+1 will be 2? right?

- jkasdhk

but 2 is not the answer

- anonymous

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

- anonymous

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

- jkasdhk

its (α^4 β^4) +1/αβ

- anonymous

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

- jkasdhk

@amrit110: How is (α^4 β^4)=1

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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