Complex mapping z ↦ ω = (z − a)/(z − b)

  • #1
Hill
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Homework Statement
Consider the complex mapping ##z \mapsto \omega = \frac {z - a} {z - b}##. Show geometrically that if we apply this mapping to the perpendicular bisector of the line-segment joining a and b, then the image is the unit circle. In greater detail, describe the motion of ω round this circle as z travels along the line at constant speed.
Relevant Equations
geometry
All points on that line are equidistant from the points a and b. Thus, the length of ##\frac {z - a} {z - b}## is 1, i.e., the points on the unit circle.
If the angle of ##z - a## is ##\alpha##, and the angle of ##z - b## is ##\beta##, then the angle of ##\frac {z - a} {z - b}## is ##\alpha - \beta##. If the point z is far toward either end of the line, ##\alpha - \beta## is close to zero. If z is in the midpoint between a and b, ##\alpha - \beta = \pi##. Thus, as z travels along the line from one "end" to another, w starts near 1, moves along one half of the unit circle speeding up towards -1, and then slows down again as it continues on the other half of the unit circle toward 1.
Are there other geometric details that I've missed?
 
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This is good, assuming that you have not already studied these transformations before and are supposed to know established theorems for them. They are called "bilinear transformations" or "Mobius transformations". I can not think of anything that you have missed.

PS There is a lot more that can be said about these transformations, but I don't think they will help to prove what you proved.
 
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  • #3
FactChecker said:
This is good, assuming that you have not already studied these transformations before and are supposed to know established theorems for them. They are called "bilinear transformations" or "Mobius transformations". I can not think of anything that you have missed.

PS There is a lot more that can be said about these transformations, but I don't think they will help to prove what you proved.
Thank you. I see that the next chapter in the textbook is "Möbius Transformations and Inversion."
 
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  • #6
I personally liked and was impressed by your proof. I think it is the right kind of reasoning to understand analytic functions of a complex variable. Of course, if you have theorems that can be used, that is the first thing to look for. But without applicable theorems, your kind of analysis is good.
 
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1. What is the complex mapping z ↦ ω = (z − a)/(z − b) used for?

The complex mapping z ↦ ω = (z − a)/(z − b) is commonly used in complex analysis to study the behavior of functions and transformations on the complex plane. It allows us to map points in the complex plane to other points in a way that preserves certain properties of the original function.

2. How does the complex mapping z ↦ ω = (z − a)/(z − b) affect the geometry of the complex plane?

The complex mapping z ↦ ω = (z − a)/(z − b) can stretch, shrink, rotate, or reflect the complex plane depending on the values of a and b. It can also map lines, circles, and other geometric shapes to different curves or regions in the complex plane.

3. What are the critical points of the complex mapping z ↦ ω = (z − a)/(z − b)?

The critical points of the complex mapping z ↦ ω = (z − a)/(z − b) are the points where the denominator (z − b) equals zero. These critical points can have a significant impact on the behavior of the mapping, such as causing singularities or discontinuities.

4. How can I visualize the complex mapping z ↦ ω = (z − a)/(z − b) on the complex plane?

One common way to visualize the complex mapping z ↦ ω = (z − a)/(z − b) is to plot a grid of points in the complex plane and map each point through the function. This can help you see how the mapping distorts or transforms the original grid of points.

5. Are there any limitations or restrictions when using the complex mapping z ↦ ω = (z − a)/(z − b)?

One limitation of the complex mapping z ↦ ω = (z − a)/(z − b) is that it may introduce singularities or discontinuities in certain regions of the complex plane, especially near the critical points where the denominator is zero. It is important to be aware of these limitations when using this mapping in complex analysis.

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