Question

已知z w 为复数,（1＋3i ）＊z 为纯虚数,w ＝z /（2＋i ）,且w 的绝对值＝5倍的跟号2,求w

Answer 1

设(1＋3i)z=bi,则z=(bi)/(1＋3i),w=z/(2＋i)=(bi)/[(1＋3i)(2＋i)],|w|=|(bi)/[(1＋3i)(2＋i)]|=|b|/[√10×√5]=5√2,解得|b|=50,则：w=(bi)/[(1＋3i)(2＋i)]=±(1－i)