`z^2+z+1=0 , z in CC`. Find `abs(sum_(k=1)^15(z^k+1/z^k)^2)=`

If z^(2)+z+1=0, then sum_(r=1)^(6)(z^(r)+z^(-r))^(2) is

If z_(k)=e^(i theta) for k=1,2,3,4, where i^(2)=-1, and if |sum_(k=1)^(4)(1)/(z_(k))|=1, then |sum_(k=1)^(4)z_(k)| is equal to

Let z_(1),z_(2),z_(3),z_(4),z_(5) and z_(6) are complex numbers lying on a unit circle with centre (0,0) . If omega=(sum_(k=1)^(6)z_(k))(sum_(k=1)^(6)(1)/(z_(k))), then

If k gt 0, k ne 1, and z_(1), z_(2) in C , then |(z-z_(1))/(z-z_(2))| = k represents

Let z_(1)z_(2),z_(3)...z_(n) be the complex numbers such that |z_(1)|=|z_(2)|...|z_(n)|=1. If z=(sum_(k=1)^(n)z_(k))(sum_(k=1)^(n)(1)/(z_(k))) then proves that z is a real number 0

Let z_(1),z_(2)..........z_(n) be equi-modular non-zero complex numbers such that z_(1)+z_(2)+z_(3)+...+z_(n)=0 Then Re(sum_(j=1)^(n)sum_(k=1)^(n)((z_(j))/(z_(k))))

Let theta_1 , theta_2 , … , theta_(10) be positive valued angles (in radian) such that theta_1+theta_2+ … +theta_(10)=2pi . Define the complex numbers z_1=e^(itheta_(1)) ,z_k= z_(k-1)e^(itheta_(k)) for k = 2, 3, … , 10 , where i = sqrt(−1) . Consider the statements ? and ? given below: P:abs(z_2-z_1)+abs(z_3-z_2)+ . . . +abs(z_(10)-z_9)+abs(z_1-z_(10)) le 2pi Q:abs(z_2^2-z_1^2)+abs(z_3^2-z_2^2)+ . . . +abs(z_(10)^2-z_9^2)+abs(z_1^2-z_(10)^2) le 4pi Then

Let z be a non - real complex number which satisfies the equation z^(23) = 1 . Then the value of sum_(22)^(k = 1 ) (1)/(1 + z ^( 8k) + z ^( 16k ))

If k gt 1, |z_(1)| lt k and |(k-z_(1)barz_(2))/(z_(1)-kz_(2))|=1 , then