`1+omega+omega^2+omega^3 = `

Which of the following cannot be valid assignment of probabilities for outcomes of sample Space S" "=" "{omega_1,omega_2,omega_3,omega_4,omega_5,omega_6,omega_7} Assignment omega_1 omega_2 omega_3 omega_4 omega_5 omega_6 omega_7

Let a sample space be S" "=" "{omega_1,omega_2, .... ,omega_6} . Which of the following assignments of probabilities to each outcome are valid? Outcomes omega_1 omega_2 omega_3 omega_4 omega_5 omega_6 (a) 1/6 1/6 1/6 1/6

If omega is a complex cube root of unity,show that ([1 omega omega^(2)omega omega^(2)1 omega^(2)1 omega]+[omega omega^(2)1 omega^(2)1 omega omega omega^(2)1])[1 omega omega^(2)]=[000]

If omega is cube root of unit, then find the value of determinant |(1,omega^3,omega^2), (omega^3,1,omega), (omega^2,omega,1)|.

|[1,omega,omega^2] , [omega, omega^2,1] , [omega^2,1,omega]|=0

(2 + omega + omega ^ (2)) ^ (3) + (1 + omega-omega ^ (2)) ^ (3) = (1-3 omega + omega ^ (2)) ^ (4) = 1

Z_1, Z_2, Z_3, ........... are terms of an A.P with common difference(1-i sqrt 3) and one of the term of A. P. being 1. omega_1,omega_2,omega_3.... are terms of a G.P. with common ratio (1+i sqrt3) and first term being 1

If 1, omega, omega^2 be three roots of 1, show that: (1+omega)^3-(1+omega^2)^3=0