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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

$$\left.\mathrm{1}\right){question}\:{incomplete} \\ $$$$ \\ $$

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