4

Find the image of the infinite strip $$0<y<1/(2c)$$ under the transformation $w=1/z$. Sketch the strip and its image.

Attempt: Clearly $$\dfrac{-v}{u^2+v^2}<\dfrac{1}{2c}$$ gives $$u^2 + (v+c)^2 > c^2$$ and the condition $y>0$, gives that $$v<0$$ I am having trouble drawing the image of this strip. Doesn't the equation above tell us that it consists of all points outside the circle $u^2+(v+c)^2 = c^2$? But another problem about the image of a half plane $$x<c_1$$ under the same transformation, states that the image should be the interior of a circle. I fail to see how the image in the above question as well as this one could be interior of a circle.

Thanks!

AAP
  • 737
  • I think it should not be that hard, given that the x coordinate is zero: $1/z=1/(x+iy)= (x-iy)/(x^2+y^2)$=$iy/y^2$ on the strip, since x=0 on the strip. – FBD Aug 13 '13 at 05:02

1 Answers1

1

It appears you have all the information to draw. The image is all points outside of that circle, and below the real axis. If you look at just the image of the boundary, the real axis mapped to the real axis and the horizontal line at $y=1/(2c)$ mapped to a circle, and the image is the region between those two.

Concerning another line mapping to a circle, this is a general property of a class of complex transformations called Linear Fractional Transformations, or Moebius Transformations, see http://en.wikipedia.org/wiki/M%C3%B6bius_transformation which map lines and circles to other lines and circles (or if you like, just circles to circles, treating lines as circles with the origin at infinity). In this case, your computations show quite explicitly how lines get transformed into circles.

Evan
  • 3,861