Question

Let `z_k= cos ((2 k pi)/(10))+ I sin ((2 k pi )/(10)), k = 1,2, ….9`

A. `{:(" "" "P " "Q" "R" "S),((a)" "(i)" "(ii)" "(iv)" "(iii)):}`
B. `{:(" "" "P " "Q" "R" "S),((b)" "(ii)" "(i)" "(iii)" "(iv)):}`
C. `{:(" "" "P " "Q" "R" "S),((c)" "(i)" "(ii)" "(iii)" "(iv)):}`
D. `{:(" "" "P " "Q" "R" "S),((c)" "(ii)" "(i)" "(iv)" "(iii)):}`

Answer 1

Correct Answer - C
(p) PLAN `e^(I theta),e^(ialpha)=e^(theta +alpha)`
Given `z_k=e^(i(2kr)/(10) rArr z_k.z_j=1=e^i(2pi/10)(k+j)`
`z_k is ` is 10 th root of unity `rArr bar zk ` will also be 10 th root of unity
Taking `z_j as bar z_k` we have `z_k.z_j=1` (True)
(Q) PLAN `e^(itheta)/(e^(ialpha))=e^((theta-alpha))`
`z=z_k//z_1=e^(i((2kpi)/10-(2pi)/10))=e^(ipi/5(k-1))`
For `k=2,z=e^(ipi/5)`
which is in the given set (False )
(R) PLAN
(i) `1- cos 2 theta = 2 sin^2 theta`
(ii) `sin 2 theta = 2 sin ^2 theta ` and
(i) ` cos 36 ^@= (sqrt(5 - 1))/(2)`
(ii) `cos 108 ^@ =(sqrt(5)+1)/(4)(|1-z_1||1-z_2|...|1-z_9|)/(10)`
Note `|1-z_k|=|1-cos""(2pik)/10-i sin "" (2 pik)/(10)|`
`=|2 sin"" (pik)/(k)||sin ""(pik)/(10)- i cos"" (pik)/10|=2|sin"" (pik)/10|`
Now required product is
`(2^9sin""pi/10sin""(2pi)/10sin""(3pi)/10 sin""(4pi)/(10))^2sin ""(5pi)/10 `
`=(2^9(sin""pi/10 cos ""pi/10.sin"" (2pi)/(10)cos ""(2pi)/10)^(2).1)/10`
`2^9(1/2sin""pi/5.(1)/2sin""(2pi)/5)^2`
`=(2^5(sin 36^@ sin 72^@)^2)/(10)`
`=2^5/(2^2xx10)(2 sin 36 ^@ sin 72^@)^2`
`=2^2/5(cos 36^@- cos 108 ^@)^2`
`=2^2/5[((sqrt5-1)/(4))+((sqrt(5+1))/(4))]^2=(2^2)/(5).(5)/4=1`
(S) Sum of nth roots of unity =0
`1+alpha + alpha ^2+alpha ^3 + ....+ alpha^9=0`
`1+underset(k=1)overset(9)Sigma alpha^k =0 `
`1+underset(k=1)overset(9)Sigma(cos""(2kpi)/10+isin"" (2kpi)/10)=0`
`1+underset(k=1)overset(9)Sigmacos ""(2kpi)/(10)=0`
`1-underset(k=1)overset(9)Sigmacos ""(2kpi)/(10)=2`
`(P)rarr (i),(Q)rarr(ii),(R)rarr(iii) ,(S) rarr (iv)`