Menu Close

Question-48982

Tinku Tara 0 Comments
Pin




Question Number 48982 by peter frank last updated on 30/Nov/18
Answered by tanmay.chaudhury50@gmail.com last updated on 01/Dec/18
2)((√3)/2)=sin((π/6)) (1/2)=cos((π/6)) e^(iθ) =cosθ+isinθ ←demoivers theorem e^(−iθ) =cosθ−isinθ z=(e^(i(π/6)) )^(107) +(e^(i((−π)/6)) )^(107) =e^(i((107π)/6)) +e^(−i((107π)/6)) =(cos((107π)/6)+isin((107π)/6))+(cos((107π)/6)−isin((107π)/6)) =2cos(((107π)/6))+i(0) so I_m (Z)=0
$$\left.\mathrm{2}\right)\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}={sin}\left(\frac{\pi}{\mathrm{6}}\right)\:\:\:\frac{\mathrm{1}}{\mathrm{2}}={cos}\left(\frac{\pi}{\mathrm{6}}\right) \\ $$$${e}^{{i}\theta} ={cos}\theta+{isin}\theta\:\leftarrow{demoivers}\:{theorem} \\ $$$${e}^{−{i}\theta} ={cos}\theta−{isin}\theta \\ $$$${z}=\left({e}^{{i}\frac{\pi}{\mathrm{6}}} \right)^{\mathrm{107}} +\left({e}^{{i}\frac{−\pi}{\mathrm{6}}} \right)^{\mathrm{107}} \\ $$$$={e}^{{i}\frac{\mathrm{107}\pi}{\mathrm{6}}} +{e}^{−{i}\frac{\mathrm{107}\pi}{\mathrm{6}}} \\ $$$$=\left({cos}\frac{\mathrm{107}\pi}{\mathrm{6}}+{isin}\frac{\mathrm{107}\pi}{\mathrm{6}}\right)+\left({cos}\frac{\mathrm{107}\pi}{\mathrm{6}}−{isin}\frac{\mathrm{107}\pi}{\mathrm{6}}\right) \\ $$$$=\mathrm{2}{cos}\left(\frac{\mathrm{107}\pi}{\mathrm{6}}\right)+{i}\left(\mathrm{0}\right) \\ $$$${so}\:{I}_{{m}} \left({Z}\right)=\mathrm{0} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 01/Dec/18
1)question incomplete
$$\left.\mathrm{1}\right){question}\:{incomplete} \\ $$$$ \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *