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}#