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As stated in the question I am asked to sketch or graph the area $$ \left| z - \frac{1}{2}\right|\leq\frac{1}{2} $$ Now in the $z$ plane this is all complex numbers that is less than $1/2$ away from the real number $1/2$

My problem is sketching the region in the $\omega=1/z$. I think it becomes a rectangle, but I am not able to prove it.

Any nudges or hints would be greatly appreciated.

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Note that $0$ is in the original set, and the image is unbounded. –  Jonas Meyer Oct 16 '12 at 21:07
    
Sorry for being unclear, my problem is sketching the region under the mapping $\omega = 1/z$, not in $z$. –  N3buchadnezzar Oct 16 '12 at 21:07

2 Answers 2

up vote 1 down vote accepted

Here are two methods:

  1. If you know something about Möbius transformations, then you know $\omega = 1/z$ is such a transformation, and they taken generalized circles to generalized circles (aka circles on the Riemann sphere). You can then determine the image of that region by looking at the image of the boundary, which is the circle $|z - 1/2| = 1/2$, which must be a line since it contains the origin. Manually calculating the image of the map at two points will allow you to determine which line it is. From here you can deduce what the image of the actual region is.

  2. If you don't know about Möbius transformations, then appealing to Euclidean coordinates, you can rewrite the region as $$(x - 1/2)^2 + y^2 \le 1/4.$$ Then inversion is $$(x,y) \mapsto \left( \frac{x}{x^2 + y^2}, \frac{-y}{x^2 + y^2}\right)$$ If you plug this formula into the above inequality, multiply out the denominators, and do a little bit of rearranging, you should obtain a nice formula.

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Your set is the closed disc with center at $(1/2,0)$ and radius $1/2$. So, if you apply the map $z\mapsto 1/z$ you can see that the boundary of the disc maps to the vertical line trough $1$ (since the unitary circle is fixed). Also the interior of the disc is mapped to the right side of the vertical line.

The complex inversion is the composite of the geometric inversion on the unitary circle with the complex conjugation, that is, $$z\mapsto\overline{\mathcal{I}_{\mathcal{C}}(z)}.$$

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