Answer to Question #256477 in Linear Algebra for johnny

Question

what is the square root of 2-i?

Answer 1

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

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

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

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

$k=0:$

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

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

$k=1:$

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

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