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Question Number 139501 by aliibrahim1 last updated on 28/Apr/21
Answered by mr W last updated on 28/Apr/21
(x+3i)^(100) =−1=e^((2k+1)πi) x_k +3i=e^((((2k+1)π)/(100))i) x_k =e^((((2k+1)π)/(100))i) −3i=cos (((2k+1)π)/(100))+(sin (((2k+1)π)/(100))−3)i =(√(10−6 sin (((2k+1)π)/(100)))) e^(−tan^(−1) ((3/(cos (((2k+1)π)/(100))))−tan (((2k+1)π)/(100)))i) Πx=Π_(k=0) ^(99) (√(10−6 sin (((2k+1)π)/(100)))) e^(−tan^(−1) ((3/(cos (((2k+1)π)/(100))))−tan (((2k+1)π)/(100)))i) =(Π_(k=0) ^(99) (√(10−6 sin (((2k+1)π)/(100))))) e^(−Σ_(k=0) ^(99) tan^(−1) ((3/(cos (((2k+1)π)/(100))))−tan (((2k+1)π)/(100)))i) ......
$$\left({x}+\mathrm{3}{i}\right)^{\mathrm{100}} =−\mathrm{1}={e}^{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi{i}} \\ $$$${x}_{{k}} +\mathrm{3}{i}={e}^{\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{\mathrm{100}}{i}} \\ $$$${x}_{{k}} ={e}^{\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{\mathrm{100}}{i}} −\mathrm{3}{i}=\mathrm{cos}\:\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{\mathrm{100}}+\left(\mathrm{sin}\:\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{\mathrm{100}}−\mathrm{3}\right){i} \\ $$$$\:\:\:=\sqrt{\mathrm{10}−\mathrm{6}\:\mathrm{sin}\:\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{\mathrm{100}}}\:{e}^{−\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{3}}{\mathrm{cos}\:\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{\mathrm{100}}}−\mathrm{tan}\:\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{\mathrm{100}}\right){i}} \\ $$$$\Pi{x}=\underset{{k}=\mathrm{0}} {\overset{\mathrm{99}} {\prod}}\sqrt{\mathrm{10}−\mathrm{6}\:\mathrm{sin}\:\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{\mathrm{100}}}\:{e}^{−\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{3}}{\mathrm{cos}\:\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{\mathrm{100}}}−\mathrm{tan}\:\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{\mathrm{100}}\right){i}} \\ $$$$=\left(\underset{{k}=\mathrm{0}} {\overset{\mathrm{99}} {\prod}}\sqrt{\mathrm{10}−\mathrm{6}\:\mathrm{sin}\:\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{\mathrm{100}}}\right)\:{e}^{−\underset{{k}=\mathrm{0}} {\overset{\mathrm{99}} {\sum}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{3}}{\mathrm{cos}\:\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{\mathrm{100}}}−\mathrm{tan}\:\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{\mathrm{100}}\right){i}} \\ $$$$…… \\ $$
Commented by aliibrahim1 last updated on 28/Apr/21
wooow thx sir
$${wooow}\:{thx}\:{sir} \\ $$

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