IIT JEE , AIEEE Forum - Mathematics , Algebra

suresh_goiitian

How many roots of the equation z¹⁰ - z⁵ - 992 = 0 have real part negative?plz answer fast with correct procedure to earn salutes

ankur.kkhurana

put z^5 = t
then solve it for t
and then
solve it for z^5

suresh_goiitian

ok.....plz complete the sol and find the no. of roots with the condition......

suresh_goiitian

someone plz help

manmeet2singh

ANS - 2 as there are 2 sign changes if we consider t=z⁵

pannaguma

z^10 - z^5 - 992=0
t^2 - t - 992=0
(t-32)(t+31)=0
thus t=32 or t=-31
thus z^5=32 or z^5=-31

???

nadeemoidu

z⁵ = 32 or -31

Let 1,z₁, z₂, z₃ ,z₄ be the fifth roots of unity.

So z= 2 , 2z₁, 2z₂, 2z₃ , 2z₄ or ⁵

(-31) , ⁵

(-31)z₁, ⁵

(-31)z₂, ⁵

(-31)z₃ , ⁵

(-31)z₄

It can be found that the arguments of the fifth roots of unity are 0, 2

/5 ,4

/5 , 6

/5 and 8

/5 .

Plot these points on the graph and the no. of negative roots can be found out.

rhd92781

put z^5=t n solve

t^2 - t - 992 = 0

u'll get t=32 n t=-31

so z⁵=32(cos2n

+isin2n

) (n is any integer)

so z=2(cos2n

/5+isin2n

/5)

again z⁵=-31(cos2n

+isin2n

) ( n - integer)

so z=-(31)^1/5(cos2n

/5+isin2n

/5)

dis r the 10 roots of the equation.

for the first 5 roots, real part will be -ve when cos(2npi/5) is -ve

it is 4 n=2 and n=3

For the next 5 roots, real part will be -ve when cos(2npi/5) is +ve.

it is for n=1, n=4, n=5 (we can't put n>5 as dere r only 10 roots)

so there r 5 roots in which the real part is -ve.

magiclko

the above solution is absolutely correct!!!