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

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Answer 1 · 2018-07-25T20:40:43

$5^{1/4}$

Given that

$\sqrt{2-i}$

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

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

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

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

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

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

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

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

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

$=5^{1/4}$

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