rishabhjaiswal225499
if z is the non real fifth root of unity then the value of ^(l 1+z+z^2+z^2+z^1l)=?



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rishabhjaiswal225499
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its find the value of 2^(l1+z+z^2+z^2z^1l) sry i wrote it wrong

goformit100
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is it IIT question ?


rishabhjaiswal225499
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you can use z+z^2+z^3+z^4=0 for z=fifth root of unity

goformit100
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Show your working.

hartnn
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so you know how to solve it already

rishabhjaiswal225499
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dw:1389460828114:dw
what next

rishabhjaiswal225499
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the options are 4,2,1,none

hartnn
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simplify this
1+z+z^2+z^2+z^1

hartnn
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(????)/z^2

rishabhjaiswal225499
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i did it by taking the denominator
dw:1389461042399:dw


hartnn
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z+z^2+z^3+z^4=0
is incorrect

hartnn
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that gives one root as z=0

hartnn
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z^5 =1
z^5 1^5 = 0

hartnn
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(z1)(1+z+z^2+z^3+z^4) = 0
as z not = 1
(1+z+z^2+z^3+z^4) = 0

hartnn
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so your entire numerator of the exponent goes to 0

hartnn
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did u follow ?
if yes, i think, you'll get the answer now :)

rishabhjaiswal225499
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no u didnt saw that i corrected the question its 1z+z^2+z^3+z^4
and not 1+z+z^2+z^3+z^4

hartnn
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oops


hartnn
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then i should say, there's typing error in the question
OR
the answer is NONE

rishabhjaiswal225499
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yup i also thought so thanks for your time by the way

hartnn
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because z is complex, so will be 2^(2/z)
welcome ^_^