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