设z = a+bi 则 z的共轭复数为 a-bi
由题意得：2a+2bi = 1-(a-bi)²
2a+2bi = 1-a²+2abi+b²
a²+2a-b²-1 + (2b-2ab)i = 0
所以a²+2a-b²-1 = 0 且 2b-2ab=0
故a=1 b=√2
那么z-√2i = 1+√2i-√2i = 1