Why radius of convergence of a power series can only be real












1












$begingroup$


I know by definition that the radius of convergence is
$R:=sup{|z|inmathbb{R}colonsum_{n=0}^{infty} a_n z^n text{ converges}}$



I don't understand why



$R:=sup{zinmathbb{C}colonsum_{n=0}^{infty} a_n z^n text{ converges}}$
it is not correct since $zinmathbb{C}$






real-analysis complex-numbers power-series







share|cite|improve this question











$endgroup$








  • 6




    $begingroup$
    Because the modulus of a complex number is a non-negative real number.
    $endgroup$
    – Bernard
    Jan 6 at 13:25






  • 2




    $begingroup$
    The supremum of an arbitrary set of complex numbers is not defined. To talk about suprema, you need an ordering.
    $endgroup$
    – Andrés E. Caicedo
    Jan 6 at 13:34










  • $begingroup$
    I have to say I think the people voting to close this for lack of context are being deceived by its brevity. I think the first sentence gives sufficient context to understand the motivation for this question, and the confusion on part of the asker.
    $endgroup$
    – jgon
    Jan 10 at 1:34
















1












$begingroup$


I know by definition that the radius of convergence is
$R:=sup{|z|inmathbb{R}colonsum_{n=0}^{infty} a_n z^n text{ converges}}$



I don't understand why



$R:=sup{zinmathbb{C}colonsum_{n=0}^{infty} a_n z^n text{ converges}}$
it is not correct since $zinmathbb{C}$






real-analysis complex-numbers power-series







share|cite|improve this question











$endgroup$








  • 6




    $begingroup$
    Because the modulus of a complex number is a non-negative real number.
    $endgroup$
    – Bernard
    Jan 6 at 13:25






  • 2




    $begingroup$
    The supremum of an arbitrary set of complex numbers is not defined. To talk about suprema, you need an ordering.
    $endgroup$
    – Andrés E. Caicedo
    Jan 6 at 13:34










  • $begingroup$
    I have to say I think the people voting to close this for lack of context are being deceived by its brevity. I think the first sentence gives sufficient context to understand the motivation for this question, and the confusion on part of the asker.
    $endgroup$
    – jgon
    Jan 10 at 1:34














1












1








1





$begingroup$


I know by definition that the radius of convergence is
$R:=sup{|z|inmathbb{R}colonsum_{n=0}^{infty} a_n z^n text{ converges}}$



I don't understand why



$R:=sup{zinmathbb{C}colonsum_{n=0}^{infty} a_n z^n text{ converges}}$
it is not correct since $zinmathbb{C}$






real-analysis complex-numbers power-series







share|cite|improve this question











$endgroup$




I know by definition that the radius of convergence is
$R:=sup{|z|inmathbb{R}colonsum_{n=0}^{infty} a_n z^n text{ converges}}$



I don't understand why



$R:=sup{zinmathbb{C}colonsum_{n=0}^{infty} a_n z^n text{ converges}}$
it is not correct since $zinmathbb{C}$






real-analysis complex-numbers power-series




real-analysis complex-numbers power-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 6 at 15:49









José Carlos Santos

160k22127232




160k22127232










asked Jan 6 at 13:23









ArchimedessArchimedess

236




236








  • 6




    $begingroup$
    Because the modulus of a complex number is a non-negative real number.
    $endgroup$
    – Bernard
    Jan 6 at 13:25






  • 2




    $begingroup$
    The supremum of an arbitrary set of complex numbers is not defined. To talk about suprema, you need an ordering.
    $endgroup$
    – Andrés E. Caicedo
    Jan 6 at 13:34










  • $begingroup$
    I have to say I think the people voting to close this for lack of context are being deceived by its brevity. I think the first sentence gives sufficient context to understand the motivation for this question, and the confusion on part of the asker.
    $endgroup$
    – jgon
    Jan 10 at 1:34














  • 6




    $begingroup$
    Because the modulus of a complex number is a non-negative real number.
    $endgroup$
    – Bernard
    Jan 6 at 13:25






  • 2




    $begingroup$
    The supremum of an arbitrary set of complex numbers is not defined. To talk about suprema, you need an ordering.
    $endgroup$
    – Andrés E. Caicedo
    Jan 6 at 13:34










  • $begingroup$
    I have to say I think the people voting to close this for lack of context are being deceived by its brevity. I think the first sentence gives sufficient context to understand the motivation for this question, and the confusion on part of the asker.
    $endgroup$
    – jgon
    Jan 10 at 1:34








6




6




$begingroup$
Because the modulus of a complex number is a non-negative real number.
$endgroup$
– Bernard
Jan 6 at 13:25




$begingroup$
Because the modulus of a complex number is a non-negative real number.
$endgroup$
– Bernard
Jan 6 at 13:25




2




2




$begingroup$
The supremum of an arbitrary set of complex numbers is not defined. To talk about suprema, you need an ordering.
$endgroup$
– Andrés E. Caicedo
Jan 6 at 13:34




$begingroup$
The supremum of an arbitrary set of complex numbers is not defined. To talk about suprema, you need an ordering.
$endgroup$
– Andrés E. Caicedo
Jan 6 at 13:34












$begingroup$
I have to say I think the people voting to close this for lack of context are being deceived by its brevity. I think the first sentence gives sufficient context to understand the motivation for this question, and the confusion on part of the asker.
$endgroup$
– jgon
Jan 10 at 1:34




$begingroup$
I have to say I think the people voting to close this for lack of context are being deceived by its brevity. I think the first sentence gives sufficient context to understand the motivation for this question, and the confusion on part of the asker.
$endgroup$
– jgon
Jan 10 at 1:34










2 Answers
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$begingroup$

Because the set$$left{zinmathbb{C},middle|,sum_{n=0}^infty a_nz^ntext{ converges}right}tag1$$either is ${0}$ or it contains complex non-real numbers. In the later case, since there is no order relation in $mathbb C$, it makes no sense to talk about the supremum of $(1)$.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    The supremum of a set $A subset B$ is defined in terms the ordering of the elements of $B$: for two elements $xin B$ and $yin B$ we need to be able to say whether $xleq y$.



    For real numbers, $leq$ is defined, and if $Asubset mathbb R$ then $A$ has a supremum in $mathbb R$.



    For the complex numbers, no such ordering exists and $x leq y$ has no meaning, so the supremum of a set of complex numbers has no meaning either.



    The clue, though, is in the word radius. This is a distance. $|z|$ is the distance of $z$ from the origin. In your example, $R$ is defined in terms of that—and convergence will in fact happen inside an actual circle of radiius $R$ on the complex plane.






    share|cite|improve this answer









    $endgroup$













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      2 Answers
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      2 Answers
      2






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      2












      $begingroup$

      Because the set$$left{zinmathbb{C},middle|,sum_{n=0}^infty a_nz^ntext{ converges}right}tag1$$either is ${0}$ or it contains complex non-real numbers. In the later case, since there is no order relation in $mathbb C$, it makes no sense to talk about the supremum of $(1)$.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        Because the set$$left{zinmathbb{C},middle|,sum_{n=0}^infty a_nz^ntext{ converges}right}tag1$$either is ${0}$ or it contains complex non-real numbers. In the later case, since there is no order relation in $mathbb C$, it makes no sense to talk about the supremum of $(1)$.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          Because the set$$left{zinmathbb{C},middle|,sum_{n=0}^infty a_nz^ntext{ converges}right}tag1$$either is ${0}$ or it contains complex non-real numbers. In the later case, since there is no order relation in $mathbb C$, it makes no sense to talk about the supremum of $(1)$.






          share|cite|improve this answer









          $endgroup$



          Because the set$$left{zinmathbb{C},middle|,sum_{n=0}^infty a_nz^ntext{ converges}right}tag1$$either is ${0}$ or it contains complex non-real numbers. In the later case, since there is no order relation in $mathbb C$, it makes no sense to talk about the supremum of $(1)$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 6 at 13:37









          José Carlos SantosJosé Carlos Santos

          160k22127232




          160k22127232























              1












              $begingroup$

              The supremum of a set $A subset B$ is defined in terms the ordering of the elements of $B$: for two elements $xin B$ and $yin B$ we need to be able to say whether $xleq y$.



              For real numbers, $leq$ is defined, and if $Asubset mathbb R$ then $A$ has a supremum in $mathbb R$.



              For the complex numbers, no such ordering exists and $x leq y$ has no meaning, so the supremum of a set of complex numbers has no meaning either.



              The clue, though, is in the word radius. This is a distance. $|z|$ is the distance of $z$ from the origin. In your example, $R$ is defined in terms of that—and convergence will in fact happen inside an actual circle of radiius $R$ on the complex plane.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                The supremum of a set $A subset B$ is defined in terms the ordering of the elements of $B$: for two elements $xin B$ and $yin B$ we need to be able to say whether $xleq y$.



                For real numbers, $leq$ is defined, and if $Asubset mathbb R$ then $A$ has a supremum in $mathbb R$.



                For the complex numbers, no such ordering exists and $x leq y$ has no meaning, so the supremum of a set of complex numbers has no meaning either.



                The clue, though, is in the word radius. This is a distance. $|z|$ is the distance of $z$ from the origin. In your example, $R$ is defined in terms of that—and convergence will in fact happen inside an actual circle of radiius $R$ on the complex plane.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  The supremum of a set $A subset B$ is defined in terms the ordering of the elements of $B$: for two elements $xin B$ and $yin B$ we need to be able to say whether $xleq y$.



                  For real numbers, $leq$ is defined, and if $Asubset mathbb R$ then $A$ has a supremum in $mathbb R$.



                  For the complex numbers, no such ordering exists and $x leq y$ has no meaning, so the supremum of a set of complex numbers has no meaning either.



                  The clue, though, is in the word radius. This is a distance. $|z|$ is the distance of $z$ from the origin. In your example, $R$ is defined in terms of that—and convergence will in fact happen inside an actual circle of radiius $R$ on the complex plane.






                  share|cite|improve this answer









                  $endgroup$



                  The supremum of a set $A subset B$ is defined in terms the ordering of the elements of $B$: for two elements $xin B$ and $yin B$ we need to be able to say whether $xleq y$.



                  For real numbers, $leq$ is defined, and if $Asubset mathbb R$ then $A$ has a supremum in $mathbb R$.



                  For the complex numbers, no such ordering exists and $x leq y$ has no meaning, so the supremum of a set of complex numbers has no meaning either.



                  The clue, though, is in the word radius. This is a distance. $|z|$ is the distance of $z$ from the origin. In your example, $R$ is defined in terms of that—and convergence will in fact happen inside an actual circle of radiius $R$ on the complex plane.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 6 at 15:38









                  timtfjtimtfj

                  2,278420




                  2,278420






























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