For `z_(1)= root6((1-i)//(1+isqrt3)), z_(2)= root6((1-i)//(sqrt3+i)), z_(3)=root6((1+i)//(sqrt3-i))` prove that `|z_(1)|= |z_(2)|= |z_(3)|`

If z_(1) = sqrt(2)(cos""(pi)/(4) + "" i sin""(pi)/(4)) and z_(2) = sqrt(3)(cos""(pi)/(3) + i sin""(pi)/(3)) , then |z_(1)z_(2)| is

If z (2- 2 sqrt3i)^(2)= i(sqrt3+i)^(4) , then arg(z)=

Let z_(1)=(2sqrt(3)+i6sqrt(7))/(6sqrt(7)+i2sqrt(3))andz_(2)=(sqrt(11)+i3sqrt(13))/(3sqrt(13)-isqrt(11)) Then, |(1)/(z_(1))+(1)/(z_(2))| is equal to a)47 b)264 c) |z_(1)-z_(2)| d) |z_(1)+z_(2)|

If z _(1) = 2 sqrt2 (1 + i) and z _(2) = 1 + isqrt3, then z _(1) ^(2) z _(2) ^(3) is equal to

Let z _(1) = 1 + i sqrt3 and z _(2) = 1 + i, then arg ((z _(1))/( z _(2))) is

If z=(pi)/(4) (1+i)^(4) ((1- sqrtpi i)/(sqrtpi + i) + (sqrtpi -i)/(1+ sqrtpi i)) , then ((|z|)/("arg"(z))) equals

If z=(-1)/(2)+i(sqrt(3))/(2) , then 8+10z+7z^(2) is equal to a) -(1)/(2)-i(sqrt(3))/(2) b) (1)/(2)+isqrt(3)/(2) c) -(1)/(2)+i(3sqrt(3))/(2) d) (sqrt(3))/(2)i

If (2 z _(1))/(3z _(2)) is a purely imaginary, then | ( z _(1) - z _(2))/( z _(1) + z _(2))| is: