Question

证明在复数范围内，方程|z|²+（1-i）-(1+i)z= (i为虚数单位)无解.

Answer 1

证明：原方程化简为|z|²+(1-i)-(1+i)z=1-3i.?

设z=x+yi(x,y∈R),代入上述方程，得x²+y²-2xi-2yi=1-3i.?

∴

将②代入①，整理得8x²-12x+5=0.                                               (*)?

∵Δ=-16＜0,

∴方程（*）无实数解.?

∴原方程在复数范围内无解.