Question

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

Answer 1

`5^{1/4}`

Explanation:

Given that

`\sqrt{2-i}`

`=(2-i)^{1/2}`

`=(\sqrt5e^{-i\tan^{-1}(1/2)})^{1/2}`

`=(\sqrt5)^{1/2}(e^{-i\tan^{-1}(1/2)})^{1/2}`

`=5^{1/4}e^{-i1/2\tan^{-1}(1/2)}`

`=5^{1/4}(\cos(1/2\tan^{-1}(1/2))+i\sin(1/2\tan^{-1}(1/2)))`

`=5^{1/4}(\sqrt{\frac{\sqrt5+2}{2\sqrt5}}+i\sqrt{\frac{\sqrt5-2}{2\sqrt5}})`

`\therefore |\sqrt{2-i}|`

`=5^{1/4}|\sqrt{\frac{\sqrt5+2}{2\sqrt5}}+i\sqrt{\frac{\sqrt5-2}{2\sqrt5}}|`

`=5^{1/4}\sqrt{\frac{\sqrt5+2}{2\sqrt5}+\frac{\sqrt5-2}{2\sqrt5}}`

`=5^{1/4}`