负数z=(1-i)/(1+i)的平方根为?
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负数z=(1-i)/(1+i)的平方根为?
负数z=(1-i)/(1+i)的平方根为?
负数z=(1-i)/(1+i)的平方根为?
z=√[(1-i)/(1+i)]
=√[(1-i)(1-i)/(1+i)(1-i)]
=√[(1-i)^2/2]
=(√2/2)(1-i)
(1-i)/(1+i)=(1-i)^2/((1+i)*(1-i))=(1-2i+i^2)/2=-i所以Z=(-i)^2=1