复数Z=(1-i)/(根号3/2+1/2i)的模

Z=(1-i)/(√3/2+1/2i)
=(1-i)(√3/2-1/2i)/(3/4+1/4)
=(1-i)(√3/2-1/2i)
=√3/2-1/2i-√3/2i+1/2i²
=(√3/2-1/2)-(√3/2+1/2)i
|Z|
=√[(√3/2-1/2)²+(√3/2+1/2)²]
=√(3/4-√3/2+1/4+3/4+√3/2+1/4)
=√2