解:Z1=R1+jXL=30+j40(Ω)=50∠53.13°。Z2=R2-jXc=80-j60(Ω)=100∠36.87°。所以电路的复导纳为:Y=1/Z1+1/Z2=1/50∠53.13°+1/100∠36.87°=0.02∠-53.13°+0.01∠-36.87°=0.012-j0.016+0.008+j0.006=0.02-j0.01(S)=0.01√5∠-26.57°。电路的阻抗为:Z=1/Y=1/0.01√5∠-26.57°=20√5∠26.57°=40+j20(Ω)。