if 1,a1,a2 ......an-1 are n roots of unity then the v alue of (1-a1)(1-a2)...(1-an-1) is
1the answer should be n..
if this a subjective question,, i dunno.. have to think n say...
but if this an objective one..
take n=3.. then u know dat, 1 +ω + ω2=0 and ω3=1.
now solve (1-ω)(1-ω2) = 3
so generalising the whole thing...
(1-a1)(1-a2)...(1-an-1)= n.
btw, wats the ans??
62Hint:
(1-a1)(1-a2)...(1-an-1)(1-an) = 1-xn
bcos
ar's are the root of 1-xn
Now you are sort of 2 steps from the final solution
1answer is n dear bt i request both of u to make ur statement more clear
62(x-a1)(x-a2)...(x-an-1)(x-an) = xn-1
bcos x=ar are the nth roots of 1
If you can agree to this..
ar = 1
then
dividbe both sides by (x-an)
we get
(x-a1)(x-a2)...(x-an-1) = (xn-1)/(x-1) = (1+x+x2+........ xn-1)
substitute x=1
The solution before this one that i gave was fundamentally wrong.. only realised half an hour later.. i wonder why none of you pointed out! :O