Correct Answer - 7
We have `z^(15) = 1`
`z=1^(1)/(15) = e^(i(2rpi)/(15)), r = 0, 1,2,.....,14`
or `z =e^(pmi(2rpi)/(15)) , r=0,1,2,....7`
The complex numbers having their arguments between `-(pi)/(2)` and `(pi)/(2)` are `e^(0), e^(pmi(2pi)/(15)),e^(pmi(4pi)/(15)),e^(pmi(6pi)/(15))`.
Hence number of solutions is7.