设复数z=a+bi,则其共轭复数z'=a-bi(1),iz'+2z=3i即i*(a-bi)+2(a+bi)=3i化简得2a+b+(a+2b-3)i=0可知2a+b=0,且a+2b-3=0解得a=-1,b=2所以z=-1+2i(2),2z+｜z'｜=-1+8i即2(a+bi)+｜a-bi｜=-1+8i即2a+根号【a^2+b^2】+2bi=...