Question Number 138879 by peter frank last updated on 19/Apr/21
Answered by Dwaipayan Shikari last updated on 19/Apr/21
$${e}^{{i}\theta\left(\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+..{n}\right)} =\mathrm{1}\Rightarrow{e}^{{i}\theta\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}} ={e}^{\mathrm{2}\pi{im}} \:\:\left({m}\in\mathbb{Z}\right) \\ $$$$\Rightarrow\theta=\frac{\mathrm{4}\pi{m}}{{n}\left({n}+\mathrm{1}\right)} \\ $$
Commented by peter frank last updated on 20/Apr/21
$${thank}\:{you}. \\ $$
Commented by peter frank last updated on 20/Apr/21
$${how}\:{part}\:{b} \\ $$