z=(√2+i)/(1+i)z=½(√2+i)(1-i)z=½[(√2+1)-(√2-1)i]答案为C
z(1+i)=|1-i|+iz(1+i)=√2 +iz(1+i)(1-i)=(√2 +i)(1-i)2z=√2 -√2i +i-1z=(√2 -1)/2 +(1-√2)i/2故z的虚部是(1-√2)/2
