z=a+bia,b∈R则a²+2abi-b²=3-4i所以a²-b²=32ab=-4b=-2/aa²-4/a²=3a^4-3a²-4=0(a²-4)(a²+1)=0所以a²=4a=±2b=-2/a所以z=2-iz=-2+i