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}#