Question and Answers Forum

All Questions      Topic List

Trigonometry Questions

Previous in All Question      Next in All Question      

Previous in Trigonometry      Next in Trigonometry      

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

Answered by JDamian last updated on 10/Apr/23

is (√(−(√(3+i)))) or (√(−(√3)+i ))?

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

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))

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com