1）已知Z1=1-2i,Z2=3+4i,求满足1/Z=(1/Z1)+(1/Z2)的复数Z2）已知Z,W为复数,（1+3i)*Z为纯虚数,W=Z/(2+i),且|W|=5根号2.求复数W

Question

1）已知Z1=1-2i,Z2=3+4i,求满足1/Z =(1/Z1)+(1/Z2)的复数Z
2）已知Z,W为复数,（1+3i)*Z为纯虚数,W=Z/(2+i),且|W|=5根号2.求复数W

Answer 1

Z1=1-2i,则1/Z1=(1+2i)/5
Z2=3+4i,则1/Z2=(3-4i)/25
则1/Z =(1/Z1)+(1/Z2)=(8+6i)/25
Z=25/(8+6i)=(4-3i)/2
因为（1+3i)*Z为纯虚数,所以设（1+3i)*Z=ki,k≠0
则Z=ki/(1+3i)=k(3+i)/10
W=Z/(2+i)=k(3+i)/10(2+i)=k(7-i)/50
又因为|W|=5√2
所以|k(7-i)/50|=5√2
得：|k|=50,k=±50
即W=±(7-i)