jkasdhk
  • jkasdhk
If α and β are complex cube roots of unity then (α^4 β^4)+1/αβ=?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
is it 2
jkasdhk
  • jkasdhk
How?
anonymous
  • anonymous
the product of the roots αβ=1 and (α^3 β^3)=1

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.