## Image of Infinite Strip under Transformation w=1/z

Q: Find the image of the infinite strip $0 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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