复数z=(m+1)+(3m-2)i m∈R,求z的模的最小值

问题描述:

复数z=(m+1)+(3m-2)i m∈R,求z的模的最小值

z=(m+1)+(3m-2)i m∈R,
z的模=√[(m+1)²+(3m-2)²]
=√[10(m²-m+1/2)]
=√[10(m-1/2)²+5/2]
当m=1/2时,z的模取到最小值=√(5/2)=√10/2