Question-190732

Question Number 190732 by zinaung last updated on 10/Apr/23

Answered by JDamian last updated on 10/Apr/23

$${is}\:\:\sqrt{−\sqrt{\mathrm{3}+{i}}}\:\:\:\:{or}\:\:\sqrt{−\sqrt{\mathrm{3}}+{i}\:\:\:}? \\ $$

Answered by Frix last updated on 10/Apr/23

(√(−(√3)+i))=(√(2e^(i((5π)/6)) ))=(√2)e^(i((5π)/(12))) =(((√3)−1)/2)+(((√3)+1)/2)i

$$\sqrt{−\sqrt{\mathrm{3}}+\mathrm{i}}=\sqrt{\mathrm{2e}^{\mathrm{i}\frac{\mathrm{5}\pi}{\mathrm{6}}} }=\sqrt{\mathrm{2}}\mathrm{e}^{\mathrm{i}\frac{\mathrm{5}\pi}{\mathrm{12}}} =\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}+\mathrm{1}}{\mathrm{2}}\mathrm{i} \\ $$

Answered by Frix last updated on 14/Apr/23

(√(−(√(3+i))))=(√(−(√((√(10))e^(i tan^(−1) (1/3)) ))))= =(√(−((10))^(1/4) e^(i((tan^(−1) (1/3))/2)) ))= =(√(((10))^(1/4) e^(i((2π+tan^(−1) (1/3))/2)) ))= =((10))^(1/8) e^(i((2π+tan^(−1) (1/3))/4))

$$\sqrt{−\sqrt{\mathrm{3}+\mathrm{i}}}=\sqrt{−\sqrt{\sqrt{\mathrm{10}}\mathrm{e}^{\mathrm{i}\:\mathrm{tan}^{−\mathrm{1}} \:\frac{\mathrm{1}}{\mathrm{3}}} }}= \\ $$$$=\sqrt{−\sqrt[{\mathrm{4}}]{\mathrm{10}}\mathrm{e}^{\mathrm{i}\frac{\mathrm{tan}^{−\mathrm{1}} \:\frac{\mathrm{1}}{\mathrm{3}}}{\mathrm{2}}} }= \\ $$$$=\sqrt{\sqrt[{\mathrm{4}}]{\mathrm{10}}\mathrm{e}^{\mathrm{i}\frac{\mathrm{2}\pi+\mathrm{tan}^{−\mathrm{1}} \:\frac{\mathrm{1}}{\mathrm{3}}}{\mathrm{2}}} }= \\ $$$$=\sqrt[{\mathrm{8}}]{\mathrm{10}}\mathrm{e}^{\mathrm{i}\frac{\mathrm{2}\pi+\mathrm{tan}^{−\mathrm{1}} \:\frac{\mathrm{1}}{\mathrm{3}}}{\mathrm{4}}} \\ $$

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