函数w=1/z,把z平面上x^2+(y-1)^2=4映射成w平面上怎样的曲线?z=1/w=1/(u+iv)=(u-iv)/(u^2+v^2)z=x+iyx=u/(u^2+v^2),y=-v/(u^2+v^2),(u/(u^2+v^2))^2+(-v/(u^2+v^2)-1)^2=4所求曲线方程为：u^2+(u^2+v^2+v)^2=4(u^2+v^2)^2其中u^2+(u^2+v^2+v)

Question

函数w=1/z,把z平面上x^2+(y-1)^2=4映射成w平面上怎样的曲线?
z=1/w=1/(u+iv)=(u-iv)/(u^2+v^2)
z=x+iy
x=u/(u^2+v^2),
y=-v/(u^2+v^2),
(u/(u^2+v^2))^2+(-v/(u^2+v^2)-1)^2=4
所求曲线方程为：u^2+(u^2+v^2+v)^2=4(u^2+v^2)^2
其中u^2+(u^2+v^2+v)^2=4(u^2+v^2)^2何解?

Answer 1

z=1/w=1/(u+iv)=(u-iv)/(u^2+v^2)
z=x+iy
x=u/(u^2+v^2),.u^2+v^2>0
y=-v/(u^2+v^2),
(u/(u^2+v^2))^2+(-v/(u^2+v^2)-1)^2=4
u^2+(u^2+v^2+v)^2=4(u^2+v^2)^2
u^2+v^2+2v(u^2+v^2)+(u^2+v^2)^2=4(u^2+v^2)^2
(1+2v)(u^2+v^2)=3(u^2+v^2)^2
1+2v=3(u^2+v^2)
u^2+(v-1/3)^2=4/9
是w平面上以（0,1/3）为圆心,2/3为半径的圆