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Question Number 91452 by  M±th+et+s last updated on 30/Apr/20
   ((−1))^(1/4)  =?
$$\:\:\:\sqrt[{\mathrm{4}}]{−\mathrm{1}}\:=? \\ $$
Commented by  M±th+et+s last updated on 30/Apr/20
thanx for solutions
$${thanx}\:{for}\:{solutions}\: \\ $$
Answered by mr W last updated on 30/Apr/20
((−1))^(1/4) =z  z^4 =−1=cos π+i sin π  z=cos (((kπ)/2)+(π/4))+i sin (((kπ)/2)+(π/4))  (k=0,1,2,3)  =((√2)/2)+((√2)/2)i  or  =−((√2)/2)+((√2)/2)i  or  =+((√2)/2)−((√2)/2)i  or  =−((√2)/2)−((√2)/2)i
$$\sqrt[{\mathrm{4}}]{−\mathrm{1}}={z} \\ $$$${z}^{\mathrm{4}} =−\mathrm{1}=\mathrm{cos}\:\pi+{i}\:\mathrm{sin}\:\pi \\ $$$${z}=\mathrm{cos}\:\left(\frac{{k}\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)+{i}\:\mathrm{sin}\:\left(\frac{{k}\pi}{\mathrm{2}}+\frac{\pi}{\mathrm{4}}\right)\:\:\left({k}=\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3}\right) \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}{i} \\ $$$${or} \\ $$$$=−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}{i} \\ $$$${or} \\ $$$$=+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}{i} \\ $$$${or} \\ $$$$=−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}{i} \\ $$
Answered by MJS last updated on 30/Apr/20
−1=e^(iπ)   (e^(iπ) )^(1/4) =e^(i(π/4)) =((√2)/2)+((√2)/2)i
$$−\mathrm{1}=\mathrm{e}^{\mathrm{i}\pi} \\ $$$$\sqrt[{\mathrm{4}}]{\mathrm{e}^{\mathrm{i}\pi} }=\mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{4}}} =\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{i} \\ $$

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