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


Submit Your Answer

please login to submit your answer

Login Here