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Question Number 58135 by rahul 19 last updated on 18/Apr/19

Answered by tanmay last updated on 18/Apr/19

$$e^{i\alpha }=cos\alpha +isin\alpha$$$$e^{i\theta }\times e^{i2\theta }\times e^{i3\theta }\times ...e^{in\theta }=1=cos(2m\pi +isin2m\pi )=e^{i\left(2m\pi \right)}$$$$e^{i\theta \left[\frac{n(n+1)}{2}\right]}=e^{i\left(2m\pi \right)}$$$$\theta =\frac{4m\pi }{n(n+1)}$$

Commented by rahul 19 last updated on 18/Apr/19

thanks sir!

Commented by tanmay last updated on 18/Apr/19

$$welcomerahul...$$

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