Question #256477
what is the square root of 2-i?
1
Expert's answer
2021-10-26T08:56:17-0400

The polar form of 2i2-i is 5(cos(tan1(0.5))+isin(tan1(0.5)))\sqrt{5}\bigg(\cos\big(-\tan^{-1}(0.5)\big)+i\sin\big(-\tan^{-1}(0.5)\big)\bigg).

According to the De Moivre's Formula all square roots of complex number 2i2-i are given by


5(cos(tan1(0.5)+2πk2)\sqrt{\sqrt{5}}\bigg(\cos\big(\dfrac{-\tan^{-1}(0.5)+2\pi k}{2}\big)

+isin(tan1(0.5)+2πk2)),k=0,1+i\sin\big(\dfrac{-\tan^{-1}(0.5)+2\pi k}{2}\big)\bigg), k=0, 1

k=0:k=0:


5(cos(tan1(0.5)+2π(0)2)\sqrt{\sqrt{5}}\bigg(\cos\big(\dfrac{-\tan^{-1}(0.5)+2\pi (0)}{2}\big)

+isin(tan1(0.5)+2π(0)2))+i\sin\big(\dfrac{-\tan^{-1}(0.5)+2\pi (0)}{2}\big)\bigg)=54(cos(tan1(0.5)2)isin(tan1(0.5)2))=\sqrt[4]{5}\bigg(\cos\big(\dfrac{\tan^{-1}(0.5)}{2}\big)-i\sin\big(\dfrac{\tan^{-1}(0.5)}{2}\big)\bigg)


k=1:k=1:


5(cos(tan1(0.5)+2π(1)2)\sqrt{\sqrt{5}}\bigg(\cos\big(\dfrac{-\tan^{-1}(0.5)+2\pi (1)}{2}\big)

+isin(tan1(0.5)+2π(1)2))+i\sin\big(\dfrac{-\tan^{-1}(0.5)+2\pi (1)}{2}\big)\bigg)=54(cos(tan1(0.5)2)+sin(tan1(0.5)2))=\sqrt[4]{5}\bigg(-\cos\big(\dfrac{\tan^{-1}(0.5)}{2}\big)+\sin\big(\dfrac{\tan^{-1}(0.5)}{2}\big)\bigg)




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