Question
The number of roots of equations z15 = 1 and Iarg zI< π/2 is
Joshi sir comment
let z = r(cosθ+i sinθ) and it is given that z15 = 1 => r15(cosθ+i sinθ)15 = 1
or r15(cos15θ+i sin15θ) = 1+0i (Here we use Demoivre's theorem)
on comparison we get sin 15θ = 0 => 15θ = nπ here n is an integer
for n = -7 to +7 we will get 15 answers for which θ will lie between -π/2 to +π/2