How does $f:zmapsto az+b$ send circles in $mathbb{C}$ to circles in $mathbb{C}$?
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I am looking at the proof in the beginning of chapter 2 of Anderson's book Hyperbolic Geometry, however I don't understand it. In particular, he seems to assume that $overline{z_1} cdot overline{z_2}=overline{z_1z_2},$ with $z_1,z_2inmathbb{C},$ but I think I have proven that that identity is false. Is there some special case where this is true?
complex-analysis hyperbolic-geometry
asked Jan 3 at 20:58
Conan G.Conan G.
519410
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add a comment |
$begingroup$
I am looking at the proof in the beginning of chapter 2 of Anderson's book Hyperbolic Geometry, however I don't understand it. In particular, he seems to assume that $overline{z_1} cdot overline{z_2}=overline{z_1z_2},$ with $z_1,z_2inmathbb{C},$ but I think I have proven that that identity is false. Is there some special case where this is true?
complex-analysis hyperbolic-geometry
asked Jan 3 at 20:58
Conan G.Conan G.
519410
$endgroup$
3
$begingroup$
That identity is always true.
$endgroup$
– Juan Diego Rojas
Jan 3 at 20:59
5
$begingroup$
Show your counterexample.
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– krirkrirk
Jan 3 at 21:00
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Oh, I thought it wasn't. Let me recheck my computation.
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– Conan G.
Jan 3 at 21:01
$begingroup$
D'oh... thanks for pointing it out.
$endgroup$
– Conan G.
Jan 3 at 21:02
add a comment |
$begingroup$
I am looking at the proof in the beginning of chapter 2 of Anderson's book Hyperbolic Geometry, however I don't understand it. In particular, he seems to assume that $overline{z_1} cdot overline{z_2}=overline{z_1z_2},$ with $z_1,z_2inmathbb{C},$ but I think I have proven that that identity is false. Is there some special case where this is true?
complex-analysis hyperbolic-geometry
asked Jan 3 at 20:58
Conan G.Conan G.
519410
$endgroup$
I am looking at the proof in the beginning of chapter 2 of Anderson's book Hyperbolic Geometry, however I don't understand it. In particular, he seems to assume that $overline{z_1} cdot overline{z_2}=overline{z_1z_2},$ with $z_1,z_2inmathbb{C},$ but I think I have proven that that identity is false. Is there some special case where this is true?
complex-analysis hyperbolic-geometry
complex-analysis hyperbolic-geometry
asked Jan 3 at 20:58
Conan G.Conan G.
519410
asked Jan 3 at 20:58
Conan G.Conan G.
519410
asked Jan 3 at 20:58
Conan G.Conan G.
519410
asked Jan 3 at 20:58
Conan G.Conan G.
519410
asked Jan 3 at 20:58
Conan G.Conan G.
519410
519410
3
$begingroup$
That identity is always true.
$endgroup$
– Juan Diego Rojas
Jan 3 at 20:59
5
$begingroup$
Show your counterexample.
$endgroup$
– krirkrirk
Jan 3 at 21:00
$begingroup$
Oh, I thought it wasn't. Let me recheck my computation.
$endgroup$
– Conan G.
Jan 3 at 21:01
$begingroup$
D'oh... thanks for pointing it out.
$endgroup$
– Conan G.
Jan 3 at 21:02
add a comment |
3
$begingroup$
That identity is always true.
$endgroup$
– Juan Diego Rojas
Jan 3 at 20:59
5
$begingroup$
Show your counterexample.
$endgroup$
– krirkrirk
Jan 3 at 21:00
$begingroup$
Oh, I thought it wasn't. Let me recheck my computation.
$endgroup$
– Conan G.
Jan 3 at 21:01
$begingroup$
D'oh... thanks for pointing it out.
$endgroup$
– Conan G.
Jan 3 at 21:02
3
3
$begingroup$
That identity is always true.
$endgroup$
– Juan Diego Rojas
Jan 3 at 20:59
$begingroup$
That identity is always true.
$endgroup$
– Juan Diego Rojas
Jan 3 at 20:59
5
5
$begingroup$
Show your counterexample.
$endgroup$
– krirkrirk
Jan 3 at 21:00
$begingroup$
Show your counterexample.
$endgroup$
– krirkrirk
Jan 3 at 21:00
$begingroup$
Oh, I thought it wasn't. Let me recheck my computation.
$endgroup$
– Conan G.
Jan 3 at 21:01
$begingroup$
Oh, I thought it wasn't. Let me recheck my computation.
$endgroup$
– Conan G.
Jan 3 at 21:01
$begingroup$
D'oh... thanks for pointing it out.
$endgroup$
– Conan G.
Jan 3 at 21:02
$begingroup$
D'oh... thanks for pointing it out.
$endgroup$
– Conan G.
Jan 3 at 21:02
add a comment |
2 Answers
2
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oldest
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To answer the question in the post title: Yes. $f(z) = az + b$ sends circles to circles. Complex multiplication is a combination of rotation and dilation, and both of those operations preserve circles. The complex addition is translation which also preserves circles.
answered Jan 3 at 21:15
Parker Glynn-AdeyParker Glynn-Adey
634
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add a comment |
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Circle: $|z-z_0| = r$, where $z_0$ is the center and $r$ the radius.
$z'= f(z)=az+b$, $z_0'=az_0+b$;
$|z'-z_0'|=|az+b-(az_0 +b)|=|a||z-z_0|=|a|r,$
i.e. a circle with center $z_0'$, and radius $|a|r$.
answered Jan 3 at 21:46
Peter SzilasPeter Szilas
11.1k2821
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
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$begingroup$
To answer the question in the post title: Yes. $f(z) = az + b$ sends circles to circles. Complex multiplication is a combination of rotation and dilation, and both of those operations preserve circles. The complex addition is translation which also preserves circles.
answered Jan 3 at 21:15
Parker Glynn-AdeyParker Glynn-Adey
634
$endgroup$
add a comment |
$begingroup$
To answer the question in the post title: Yes. $f(z) = az + b$ sends circles to circles. Complex multiplication is a combination of rotation and dilation, and both of those operations preserve circles. The complex addition is translation which also preserves circles.
answered Jan 3 at 21:15
Parker Glynn-AdeyParker Glynn-Adey
634
$endgroup$
add a comment |
$begingroup$
To answer the question in the post title: Yes. $f(z) = az + b$ sends circles to circles. Complex multiplication is a combination of rotation and dilation, and both of those operations preserve circles. The complex addition is translation which also preserves circles.
answered Jan 3 at 21:15
Parker Glynn-AdeyParker Glynn-Adey
634
$endgroup$
To answer the question in the post title: Yes. $f(z) = az + b$ sends circles to circles. Complex multiplication is a combination of rotation and dilation, and both of those operations preserve circles. The complex addition is translation which also preserves circles.
answered Jan 3 at 21:15
Parker Glynn-AdeyParker Glynn-Adey
634
answered Jan 3 at 21:15
Parker Glynn-AdeyParker Glynn-Adey
634
answered Jan 3 at 21:15
Parker Glynn-AdeyParker Glynn-Adey
634
answered Jan 3 at 21:15
Parker Glynn-AdeyParker Glynn-Adey
634
634
add a comment |
add a comment |
$begingroup$
Circle: $|z-z_0| = r$, where $z_0$ is the center and $r$ the radius.
$z'= f(z)=az+b$, $z_0'=az_0+b$;
$|z'-z_0'|=|az+b-(az_0 +b)|=|a||z-z_0|=|a|r,$
i.e. a circle with center $z_0'$, and radius $|a|r$.
answered Jan 3 at 21:46
Peter SzilasPeter Szilas
11.1k2821
$endgroup$
add a comment |
$begingroup$
Circle: $|z-z_0| = r$, where $z_0$ is the center and $r$ the radius.
$z'= f(z)=az+b$, $z_0'=az_0+b$;
$|z'-z_0'|=|az+b-(az_0 +b)|=|a||z-z_0|=|a|r,$
i.e. a circle with center $z_0'$, and radius $|a|r$.
answered Jan 3 at 21:46
Peter SzilasPeter Szilas
11.1k2821
$endgroup$
add a comment |
$begingroup$
Circle: $|z-z_0| = r$, where $z_0$ is the center and $r$ the radius.
$z'= f(z)=az+b$, $z_0'=az_0+b$;
$|z'-z_0'|=|az+b-(az_0 +b)|=|a||z-z_0|=|a|r,$
i.e. a circle with center $z_0'$, and radius $|a|r$.
answered Jan 3 at 21:46
Peter SzilasPeter Szilas
11.1k2821
$endgroup$
Circle: $|z-z_0| = r$, where $z_0$ is the center and $r$ the radius.
$z'= f(z)=az+b$, $z_0'=az_0+b$;
$|z'-z_0'|=|az+b-(az_0 +b)|=|a||z-z_0|=|a|r,$
i.e. a circle with center $z_0'$, and radius $|a|r$.
answered Jan 3 at 21:46
Peter SzilasPeter Szilas
11.1k2821
answered Jan 3 at 21:46
Peter SzilasPeter Szilas
11.1k2821
answered Jan 3 at 21:46
Peter SzilasPeter Szilas
11.1k2821
answered Jan 3 at 21:46
Peter SzilasPeter Szilas
11.1k2821
11.1k2821
add a comment |
add a comment |
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3
$begingroup$
That identity is always true.
$endgroup$
– Juan Diego Rojas
Jan 3 at 20:59
5
$begingroup$
Show your counterexample.
$endgroup$
– krirkrirk
Jan 3 at 21:00
$begingroup$
Oh, I thought it wasn't. Let me recheck my computation.
$endgroup$
– Conan G.
Jan 3 at 21:01
$begingroup$
D'oh... thanks for pointing it out.
$endgroup$
– Conan G.
Jan 3 at 21:02