Image of Infinite Strip under Transformation w=1/z

Q: Find the image of the infinite strip 0<y<1/(2c) under the transformation w=1/z.

(This question is taken from Complex Variables and Applications (Brown and Churchill))

Answer: u^2+(v+c)^2>c^2, v<0.

Solution:

We are using the standard notation w=u+iv, z=x+iy. From the equation \displaystyle z=\frac{1}{w}=\frac{1}{u+iv}=\frac{u-iv}{u^2+v^2}, we can conclude that \displaystyle y=\frac{-v}{u^2+v^2}.

With some algebra and completing the square, one can obtain u^2+(v+\frac{1}{2y})^2=(\frac{1}{2y})^2, which is the equation of a circle centered at (0,-\frac{1}{2y}) with radius 1/2y. When y=1/(2c), the equation of the circle is u^2+(v+c)^2=c^2. As y gets smaller (closer to zero), the radius of the circle becomes larger, while still remaining tangent to the horizontal axis.

Thus, the image of the strip is u^2+(v+c)^2>c^2, v<0.

 

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One Response to Image of Infinite Strip under Transformation w=1/z

  1. Pingback: Image of Vertical Strip under Inversion | Singapore Maths Tuition

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