How does $f:zmapsto az+b$ send circles in $mathbb{C}$ to circles in $mathbb{C}$?












0














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?










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  • 3




    That identity is always true.
    – Juan Diego Rojas
    Jan 3 at 20:59






  • 5




    Show your counterexample.
    – krirkrirk
    Jan 3 at 21:00










  • Oh, I thought it wasn't. Let me recheck my computation.
    – Conan G.
    Jan 3 at 21:01










  • D'oh... thanks for pointing it out.
    – Conan G.
    Jan 3 at 21:02
















0














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?










share|cite|improve this question


















  • 3




    That identity is always true.
    – Juan Diego Rojas
    Jan 3 at 20:59






  • 5




    Show your counterexample.
    – krirkrirk
    Jan 3 at 21:00










  • Oh, I thought it wasn't. Let me recheck my computation.
    – Conan G.
    Jan 3 at 21:01










  • D'oh... thanks for pointing it out.
    – Conan G.
    Jan 3 at 21:02














0












0








0







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?










share|cite|improve this question













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






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asked Jan 3 at 20:58









Conan G.

519410




519410








  • 3




    That identity is always true.
    – Juan Diego Rojas
    Jan 3 at 20:59






  • 5




    Show your counterexample.
    – krirkrirk
    Jan 3 at 21:00










  • Oh, I thought it wasn't. Let me recheck my computation.
    – Conan G.
    Jan 3 at 21:01










  • D'oh... thanks for pointing it out.
    – Conan G.
    Jan 3 at 21:02














  • 3




    That identity is always true.
    – Juan Diego Rojas
    Jan 3 at 20:59






  • 5




    Show your counterexample.
    – krirkrirk
    Jan 3 at 21:00










  • Oh, I thought it wasn't. Let me recheck my computation.
    – Conan G.
    Jan 3 at 21:01










  • D'oh... thanks for pointing it out.
    – Conan G.
    Jan 3 at 21:02








3




3




That identity is always true.
– Juan Diego Rojas
Jan 3 at 20:59




That identity is always true.
– Juan Diego Rojas
Jan 3 at 20:59




5




5




Show your counterexample.
– krirkrirk
Jan 3 at 21:00




Show your counterexample.
– krirkrirk
Jan 3 at 21:00












Oh, I thought it wasn't. Let me recheck my computation.
– Conan G.
Jan 3 at 21:01




Oh, I thought it wasn't. Let me recheck my computation.
– Conan G.
Jan 3 at 21:01












D'oh... thanks for pointing it out.
– Conan G.
Jan 3 at 21:02




D'oh... thanks for pointing it out.
– Conan G.
Jan 3 at 21:02










2 Answers
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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.






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    0














    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 $|a|r$.






    share|cite|improve this answer





















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      2 Answers
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      2 Answers
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      2














      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.






      share|cite|improve this answer


























        2














        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.






        share|cite|improve this answer
























          2












          2








          2






          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.






          share|cite|improve this answer












          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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 3 at 21:15









          Parker Glynn-Adey

          553




          553























              0














              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 $|a|r$.






              share|cite|improve this answer


























                0














                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 $|a|r$.






                share|cite|improve this answer
























                  0












                  0








                  0






                  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 $|a|r$.






                  share|cite|improve this answer












                  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 $|a|r$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 2 days ago









                  Peter Szilas

                  10.7k2720




                  10.7k2720






























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