1. ## linear mapping

Consider the inversion f(z)=1/z. What happens to the line y=2 under this mapping?

In such a question the way to solve them is:
1/z=1/(x+iy)=(x-iy)/(x^2+y^2)
so u= x/(x^2+y^2) and v=(-y)/(x^2+y^2)
so x= u/(u^2+v^2) y=-v/(u^2+v^2)

u^2 + v^2 +v/2=0 by complete the square
=> u^2+(v+1/4)^2=1/16

Thanks

2. Originally Posted by NaNa
Consider the inversion f(z)=1/z. What happens to the line y=2 under this mapping?

In such a question the way to solve them is:
1/z=1/(x+iy)=(x-iy)/(x^2+y^2)
so u= x/(x^2+y^2) and v=(-y)/(x^2+y^2)
so x= u/(u^2+v^2) y=-v/(u^2+v^2)

u^2 + v^2 +v/2=0 by complete the square
=> u^2+(v+1/4)^2=1/16

Thanks
You might want to say that the radius is 1/4 rather than 1/root(16)!

3. yes!

4. so is right!?