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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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