1.已知方程x^2+z*x+4+3i=0有实数根,求复数z的模的最小值

设z=a+bi
x^2+z^x+4+3i=x^2+ax+bxi+4+3i=0
x^2+ax+4=0 -a=x+4/x a^2=x^2+8+16/x^2
bx+3=0 -b=3/x b^2=9/x^2
a^+b^2=x^2+25/x^2+8>=2根号[x^2*25/x^2]+8=10+8=18
复数z的模|Z|=根号(a^2+b^2)>=根号18=3根号2
复数z的模的最小值3根号2