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.
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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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  • #2
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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