One of the values of ((1+i)/root2)^2/3 is (1) root3 + i (2) - i (3) i (4) -root3 + i

Solution:

Let Z = ((1+i)/√2)^2/3

= ((1/√2)+(i/√2))^2/3

Z = (cos (π/4) + i sin (π/4))^2/3

Z = cos (8k+1)π/6 + i sin (8k+1)π/6

Substitute k = 1, we get

Z = cos (8+1)π/6 + i sin (8+1)π/6

= cos 9π/6 + i sin 9π/6

= cos 3π/2 + i sin 3π/2

= 0-1(i)

= -i

Hence option (2) is the answer.