How do you find the absolute value of sqrt(2-i)?

${5}^{\frac{1}{4}}$

Explanation:

Given that

$\setminus \sqrt{2 - i}$

$= {\left(2 - i\right)}^{\frac{1}{2}}$

$= {\left(\setminus \sqrt{5} {e}^{- i \setminus {\tan}^{- 1} \left(\frac{1}{2}\right)}\right)}^{\frac{1}{2}}$

$= {\left(\setminus \sqrt{5}\right)}^{\frac{1}{2}} {\left({e}^{- i \setminus {\tan}^{- 1} \left(\frac{1}{2}\right)}\right)}^{\frac{1}{2}}$

$= {5}^{\frac{1}{4}} {e}^{- i \frac{1}{2} \setminus {\tan}^{- 1} \left(\frac{1}{2}\right)}$

$= {5}^{\frac{1}{4}} \left(\setminus \cos \left(\frac{1}{2} \setminus {\tan}^{- 1} \left(\frac{1}{2}\right)\right) + i \setminus \sin \left(\frac{1}{2} \setminus {\tan}^{- 1} \left(\frac{1}{2}\right)\right)\right)$

$= {5}^{\frac{1}{4}} \left(\setminus \sqrt{\setminus \frac{\setminus \sqrt{5} + 2}{2 \setminus \sqrt{5}}} + i \setminus \sqrt{\setminus \frac{\setminus \sqrt{5} - 2}{2 \setminus \sqrt{5}}}\right)$

$\setminus \therefore | \setminus \sqrt{2 - i} |$

$= {5}^{\frac{1}{4}} | \setminus \sqrt{\setminus \frac{\setminus \sqrt{5} + 2}{2 \setminus \sqrt{5}}} + i \setminus \sqrt{\setminus \frac{\setminus \sqrt{5} - 2}{2 \setminus \sqrt{5}}} |$

$= {5}^{\frac{1}{4}} \setminus \sqrt{\setminus \frac{\setminus \sqrt{5} + 2}{2 \setminus \sqrt{5}} + \setminus \frac{\setminus \sqrt{5} - 2}{2 \setminus \sqrt{5}}}$

$= {5}^{\frac{1}{4}}$