consider the mobius transformation $f(z) =frac{1}{z} $ , $z in mathbb{C}$, $z neq 0$.choose the correct...












0












$begingroup$


consider the mobius transformation $f(z) =frac{1}{z} $ , $z in mathbb{C}$, $z neq 0$. If C denotes a circle with positive radius passing through the origin, then $f$ map $C setminus {0}$ to



choose the correct options



$1.$Circle



$2.$ a line



$3.$ a line passing through the origin



$4.$ a line not passing through the origin



My attempt : i take $z = e^{itheta}$,$f(z) = 1/z = 1/e^{itheta}=e^{-itheta}$



that $f(z) = costheta - isintheta$



After that im not able to proceed further pliz help me



any hints/solution will be appreciated thanks u






complex-analysis complex-numbers







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    consider the mobius transformation $f(z) =frac{1}{z} $ , $z in mathbb{C}$, $z neq 0$. If C denotes a circle with positive radius passing through the origin, then $f$ map $C setminus {0}$ to



    choose the correct options



    $1.$Circle



    $2.$ a line



    $3.$ a line passing through the origin



    $4.$ a line not passing through the origin



    My attempt : i take $z = e^{itheta}$,$f(z) = 1/z = 1/e^{itheta}=e^{-itheta}$



    that $f(z) = costheta - isintheta$



    After that im not able to proceed further pliz help me



    any hints/solution will be appreciated thanks u






    complex-analysis complex-numbers







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      consider the mobius transformation $f(z) =frac{1}{z} $ , $z in mathbb{C}$, $z neq 0$. If C denotes a circle with positive radius passing through the origin, then $f$ map $C setminus {0}$ to



      choose the correct options



      $1.$Circle



      $2.$ a line



      $3.$ a line passing through the origin



      $4.$ a line not passing through the origin



      My attempt : i take $z = e^{itheta}$,$f(z) = 1/z = 1/e^{itheta}=e^{-itheta}$



      that $f(z) = costheta - isintheta$



      After that im not able to proceed further pliz help me



      any hints/solution will be appreciated thanks u






      complex-analysis complex-numbers







      share|cite|improve this question











      $endgroup$




      consider the mobius transformation $f(z) =frac{1}{z} $ , $z in mathbb{C}$, $z neq 0$. If C denotes a circle with positive radius passing through the origin, then $f$ map $C setminus {0}$ to



      choose the correct options



      $1.$Circle



      $2.$ a line



      $3.$ a line passing through the origin



      $4.$ a line not passing through the origin



      My attempt : i take $z = e^{itheta}$,$f(z) = 1/z = 1/e^{itheta}=e^{-itheta}$



      that $f(z) = costheta - isintheta$



      After that im not able to proceed further pliz help me



      any hints/solution will be appreciated thanks u






      complex-analysis complex-numbers




      complex-analysis complex-numbers






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 13 '18 at 12:30









      José Carlos Santos

      160k22126232




      160k22126232










      asked Dec 13 '18 at 12:23









      jasminejasmine

      1,706417




      1,706417






















          2 Answers
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          active

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          2












          $begingroup$

          What's the point of taking $z=e^{itheta}$ that gives you a circle not passing through the origin? If you do take a circle passing through the origin, then what you will get is a line not passing through the origin. That is, the third option is the right one.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            im not getting @Jose carlos sir ....can u elaborate more,,,,,
            $endgroup$
            – jasmine
            Dec 13 '18 at 12:35






          • 1




            $begingroup$
            Don't you get that your circle doesn't pass through the origin?
            $endgroup$
            – José Carlos Santos
            Dec 13 '18 at 12:38










          • $begingroup$
            my confusion is that why its is not map circle to circle ?
            $endgroup$
            – jasmine
            Dec 13 '18 at 12:47






          • 1




            $begingroup$
            It maps your circle into a circle, but since your circle doesn't pass through the origin, that does not mapr. Now, comput$$fbigl({e^{itheta}+1,|,thetain[0,2pi]}bigr)$$and see what you get then.
            $endgroup$
            – José Carlos Santos
            Dec 13 '18 at 13:00








          • 1




            $begingroup$
            It matches my eyes, doesn't it?! ;-)
            $endgroup$
            – José Carlos Santos
            Dec 13 '18 at 14:24



















          1












          $begingroup$

          You can consider the above Möbius transformation $f$ on the extended complex plane $hat{mathbb C} =mathbb C cup{ infty}$ with $f(0)= infty$ and $f(infty)=0$.



          Since $0 in C$ we have $ infty in f(hat{mathbb C})$. And since $f(infty)=0$, we see that the third option is correct.






          share|cite|improve this answer









          $endgroup$













            Your Answer





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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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            2












            $begingroup$

            What's the point of taking $z=e^{itheta}$ that gives you a circle not passing through the origin? If you do take a circle passing through the origin, then what you will get is a line not passing through the origin. That is, the third option is the right one.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              im not getting @Jose carlos sir ....can u elaborate more,,,,,
              $endgroup$
              – jasmine
              Dec 13 '18 at 12:35






            • 1




              $begingroup$
              Don't you get that your circle doesn't pass through the origin?
              $endgroup$
              – José Carlos Santos
              Dec 13 '18 at 12:38










            • $begingroup$
              my confusion is that why its is not map circle to circle ?
              $endgroup$
              – jasmine
              Dec 13 '18 at 12:47






            • 1




              $begingroup$
              It maps your circle into a circle, but since your circle doesn't pass through the origin, that does not mapr. Now, comput$$fbigl({e^{itheta}+1,|,thetain[0,2pi]}bigr)$$and see what you get then.
              $endgroup$
              – José Carlos Santos
              Dec 13 '18 at 13:00








            • 1




              $begingroup$
              It matches my eyes, doesn't it?! ;-)
              $endgroup$
              – José Carlos Santos
              Dec 13 '18 at 14:24
















            2












            $begingroup$

            What's the point of taking $z=e^{itheta}$ that gives you a circle not passing through the origin? If you do take a circle passing through the origin, then what you will get is a line not passing through the origin. That is, the third option is the right one.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              im not getting @Jose carlos sir ....can u elaborate more,,,,,
              $endgroup$
              – jasmine
              Dec 13 '18 at 12:35






            • 1




              $begingroup$
              Don't you get that your circle doesn't pass through the origin?
              $endgroup$
              – José Carlos Santos
              Dec 13 '18 at 12:38










            • $begingroup$
              my confusion is that why its is not map circle to circle ?
              $endgroup$
              – jasmine
              Dec 13 '18 at 12:47






            • 1




              $begingroup$
              It maps your circle into a circle, but since your circle doesn't pass through the origin, that does not mapr. Now, comput$$fbigl({e^{itheta}+1,|,thetain[0,2pi]}bigr)$$and see what you get then.
              $endgroup$
              – José Carlos Santos
              Dec 13 '18 at 13:00








            • 1




              $begingroup$
              It matches my eyes, doesn't it?! ;-)
              $endgroup$
              – José Carlos Santos
              Dec 13 '18 at 14:24














            2












            2








            2





            $begingroup$

            What's the point of taking $z=e^{itheta}$ that gives you a circle not passing through the origin? If you do take a circle passing through the origin, then what you will get is a line not passing through the origin. That is, the third option is the right one.






            share|cite|improve this answer









            $endgroup$



            What's the point of taking $z=e^{itheta}$ that gives you a circle not passing through the origin? If you do take a circle passing through the origin, then what you will get is a line not passing through the origin. That is, the third option is the right one.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 13 '18 at 12:29









            José Carlos SantosJosé Carlos Santos

            160k22126232




            160k22126232












            • $begingroup$
              im not getting @Jose carlos sir ....can u elaborate more,,,,,
              $endgroup$
              – jasmine
              Dec 13 '18 at 12:35






            • 1




              $begingroup$
              Don't you get that your circle doesn't pass through the origin?
              $endgroup$
              – José Carlos Santos
              Dec 13 '18 at 12:38










            • $begingroup$
              my confusion is that why its is not map circle to circle ?
              $endgroup$
              – jasmine
              Dec 13 '18 at 12:47






            • 1




              $begingroup$
              It maps your circle into a circle, but since your circle doesn't pass through the origin, that does not mapr. Now, comput$$fbigl({e^{itheta}+1,|,thetain[0,2pi]}bigr)$$and see what you get then.
              $endgroup$
              – José Carlos Santos
              Dec 13 '18 at 13:00








            • 1




              $begingroup$
              It matches my eyes, doesn't it?! ;-)
              $endgroup$
              – José Carlos Santos
              Dec 13 '18 at 14:24


















            • $begingroup$
              im not getting @Jose carlos sir ....can u elaborate more,,,,,
              $endgroup$
              – jasmine
              Dec 13 '18 at 12:35






            • 1




              $begingroup$
              Don't you get that your circle doesn't pass through the origin?
              $endgroup$
              – José Carlos Santos
              Dec 13 '18 at 12:38










            • $begingroup$
              my confusion is that why its is not map circle to circle ?
              $endgroup$
              – jasmine
              Dec 13 '18 at 12:47






            • 1




              $begingroup$
              It maps your circle into a circle, but since your circle doesn't pass through the origin, that does not mapr. Now, comput$$fbigl({e^{itheta}+1,|,thetain[0,2pi]}bigr)$$and see what you get then.
              $endgroup$
              – José Carlos Santos
              Dec 13 '18 at 13:00








            • 1




              $begingroup$
              It matches my eyes, doesn't it?! ;-)
              $endgroup$
              – José Carlos Santos
              Dec 13 '18 at 14:24
















            $begingroup$
            im not getting @Jose carlos sir ....can u elaborate more,,,,,
            $endgroup$
            – jasmine
            Dec 13 '18 at 12:35




            $begingroup$
            im not getting @Jose carlos sir ....can u elaborate more,,,,,
            $endgroup$
            – jasmine
            Dec 13 '18 at 12:35




            1




            1




            $begingroup$
            Don't you get that your circle doesn't pass through the origin?
            $endgroup$
            – José Carlos Santos
            Dec 13 '18 at 12:38




            $begingroup$
            Don't you get that your circle doesn't pass through the origin?
            $endgroup$
            – José Carlos Santos
            Dec 13 '18 at 12:38












            $begingroup$
            my confusion is that why its is not map circle to circle ?
            $endgroup$
            – jasmine
            Dec 13 '18 at 12:47




            $begingroup$
            my confusion is that why its is not map circle to circle ?
            $endgroup$
            – jasmine
            Dec 13 '18 at 12:47




            1




            1




            $begingroup$
            It maps your circle into a circle, but since your circle doesn't pass through the origin, that does not mapr. Now, comput$$fbigl({e^{itheta}+1,|,thetain[0,2pi]}bigr)$$and see what you get then.
            $endgroup$
            – José Carlos Santos
            Dec 13 '18 at 13:00






            $begingroup$
            It maps your circle into a circle, but since your circle doesn't pass through the origin, that does not mapr. Now, comput$$fbigl({e^{itheta}+1,|,thetain[0,2pi]}bigr)$$and see what you get then.
            $endgroup$
            – José Carlos Santos
            Dec 13 '18 at 13:00






            1




            1




            $begingroup$
            It matches my eyes, doesn't it?! ;-)
            $endgroup$
            – José Carlos Santos
            Dec 13 '18 at 14:24




            $begingroup$
            It matches my eyes, doesn't it?! ;-)
            $endgroup$
            – José Carlos Santos
            Dec 13 '18 at 14:24











            1












            $begingroup$

            You can consider the above Möbius transformation $f$ on the extended complex plane $hat{mathbb C} =mathbb C cup{ infty}$ with $f(0)= infty$ and $f(infty)=0$.



            Since $0 in C$ we have $ infty in f(hat{mathbb C})$. And since $f(infty)=0$, we see that the third option is correct.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              You can consider the above Möbius transformation $f$ on the extended complex plane $hat{mathbb C} =mathbb C cup{ infty}$ with $f(0)= infty$ and $f(infty)=0$.



              Since $0 in C$ we have $ infty in f(hat{mathbb C})$. And since $f(infty)=0$, we see that the third option is correct.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                You can consider the above Möbius transformation $f$ on the extended complex plane $hat{mathbb C} =mathbb C cup{ infty}$ with $f(0)= infty$ and $f(infty)=0$.



                Since $0 in C$ we have $ infty in f(hat{mathbb C})$. And since $f(infty)=0$, we see that the third option is correct.






                share|cite|improve this answer









                $endgroup$



                You can consider the above Möbius transformation $f$ on the extended complex plane $hat{mathbb C} =mathbb C cup{ infty}$ with $f(0)= infty$ and $f(infty)=0$.



                Since $0 in C$ we have $ infty in f(hat{mathbb C})$. And since $f(infty)=0$, we see that the third option is correct.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 13 '18 at 12:45









                FredFred

                46k1848




                46k1848






























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