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Question Number 139027 by mohammad17 last updated on 21/Apr/21
find Re(z) and Im(z) of [(e^(i(1+2k)π) )^(1/(10)) +1]^(−1)
$${find}\:{Re}\left({z}\right)\:{and}\:{Im}\left({z}\right)\:{of}\:\left[\left({e}^{{i}\left(\mathrm{1}+\mathrm{2}{k}\right)\pi} \right)^{\frac{\mathrm{1}}{\mathrm{10}}} +\mathrm{1}\right]^{−\mathrm{1}} \\ $$$$ \\ $$
Commented by mohammad17 last updated on 21/Apr/21
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Answered by MJS_new last updated on 21/Apr/21
e^(i(1+2k)π) =−1  (−1)^(1/(10)) =e^((iπ)/(10)) =((√(10+2(√5)))/4)−((1−(√5))/4)i  (1/(1+((√(10+2(√5)))/4)−((1−(√5))/4)i))=(1/2)+(((1+(√5)−(√(5+2(√5))))/2))i
$$\mathrm{e}^{\mathrm{i}\left(\mathrm{1}+\mathrm{2}{k}\right)\pi} =−\mathrm{1} \\ $$$$\left(−\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{10}}} =\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{10}}} =\frac{\sqrt{\mathrm{10}+\mathrm{2}\sqrt{\mathrm{5}}}}{\mathrm{4}}−\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{4}}\mathrm{i} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}+\frac{\sqrt{\mathrm{10}+\mathrm{2}\sqrt{\mathrm{5}}}}{\mathrm{4}}−\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{4}}\mathrm{i}}=\frac{\mathrm{1}}{\mathrm{2}}+\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}−\sqrt{\mathrm{5}+\mathrm{2}\sqrt{\mathrm{5}}}}{\mathrm{2}}\right)\mathrm{i} \\ $$

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