If $f$ complex polynomial such that $f(z) in mathbb{R}$ for all $|z| = 1$, then $f$ is constant.












5












$begingroup$


Let $f in mathbb{C}[z]$ a complex polynomial such that $f(mathbb{D}) subset mathbb{R},$ where
$$
mathbb{D} = { z in mathbb{C} : |z| = 1 }
$$

Show that $f$ is constant.



My attempt is: define a function $g(z) = e^{if(z)}$ and plug some roots of unity in $g,$ but how can I conclude from this?
I know that if $Omega$ is open, connected and $f(Omega) subset mathbb{R},$ then $f$ is constant.










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$endgroup$












  • $begingroup$
    Can you can break $mathbb{D}$ into two open charts and conclude that if it's constant on those charts and where the charts overlap then it must be constant on the entire set? I want to puncture the circle at a single point, conclude it's constant on that open set, then do it again with a different point and then conclude it's constant everywhere.
    $endgroup$
    – CyclotomicField
    Nov 28 '18 at 18:30










  • $begingroup$
    The other answers are slick (and correct), but I think I have a from scratch proof that might be what you are looking for.
    $endgroup$
    – Matematleta
    Nov 28 '18 at 22:33
















5












$begingroup$


Let $f in mathbb{C}[z]$ a complex polynomial such that $f(mathbb{D}) subset mathbb{R},$ where
$$
mathbb{D} = { z in mathbb{C} : |z| = 1 }
$$

Show that $f$ is constant.



My attempt is: define a function $g(z) = e^{if(z)}$ and plug some roots of unity in $g,$ but how can I conclude from this?
I know that if $Omega$ is open, connected and $f(Omega) subset mathbb{R},$ then $f$ is constant.










share|cite|improve this question











$endgroup$












  • $begingroup$
    Can you can break $mathbb{D}$ into two open charts and conclude that if it's constant on those charts and where the charts overlap then it must be constant on the entire set? I want to puncture the circle at a single point, conclude it's constant on that open set, then do it again with a different point and then conclude it's constant everywhere.
    $endgroup$
    – CyclotomicField
    Nov 28 '18 at 18:30










  • $begingroup$
    The other answers are slick (and correct), but I think I have a from scratch proof that might be what you are looking for.
    $endgroup$
    – Matematleta
    Nov 28 '18 at 22:33














5












5








5


2



$begingroup$


Let $f in mathbb{C}[z]$ a complex polynomial such that $f(mathbb{D}) subset mathbb{R},$ where
$$
mathbb{D} = { z in mathbb{C} : |z| = 1 }
$$

Show that $f$ is constant.



My attempt is: define a function $g(z) = e^{if(z)}$ and plug some roots of unity in $g,$ but how can I conclude from this?
I know that if $Omega$ is open, connected and $f(Omega) subset mathbb{R},$ then $f$ is constant.










share|cite|improve this question











$endgroup$




Let $f in mathbb{C}[z]$ a complex polynomial such that $f(mathbb{D}) subset mathbb{R},$ where
$$
mathbb{D} = { z in mathbb{C} : |z| = 1 }
$$

Show that $f$ is constant.



My attempt is: define a function $g(z) = e^{if(z)}$ and plug some roots of unity in $g,$ but how can I conclude from this?
I know that if $Omega$ is open, connected and $f(Omega) subset mathbb{R},$ then $f$ is constant.







complex-analysis polynomials






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share|cite|improve this question













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edited Nov 28 '18 at 18:37









Brahadeesh

6,18742361




6,18742361










asked Nov 28 '18 at 16:26









674123173797 - 4674123173797 - 4

1457




1457












  • $begingroup$
    Can you can break $mathbb{D}$ into two open charts and conclude that if it's constant on those charts and where the charts overlap then it must be constant on the entire set? I want to puncture the circle at a single point, conclude it's constant on that open set, then do it again with a different point and then conclude it's constant everywhere.
    $endgroup$
    – CyclotomicField
    Nov 28 '18 at 18:30










  • $begingroup$
    The other answers are slick (and correct), but I think I have a from scratch proof that might be what you are looking for.
    $endgroup$
    – Matematleta
    Nov 28 '18 at 22:33


















  • $begingroup$
    Can you can break $mathbb{D}$ into two open charts and conclude that if it's constant on those charts and where the charts overlap then it must be constant on the entire set? I want to puncture the circle at a single point, conclude it's constant on that open set, then do it again with a different point and then conclude it's constant everywhere.
    $endgroup$
    – CyclotomicField
    Nov 28 '18 at 18:30










  • $begingroup$
    The other answers are slick (and correct), but I think I have a from scratch proof that might be what you are looking for.
    $endgroup$
    – Matematleta
    Nov 28 '18 at 22:33
















$begingroup$
Can you can break $mathbb{D}$ into two open charts and conclude that if it's constant on those charts and where the charts overlap then it must be constant on the entire set? I want to puncture the circle at a single point, conclude it's constant on that open set, then do it again with a different point and then conclude it's constant everywhere.
$endgroup$
– CyclotomicField
Nov 28 '18 at 18:30




$begingroup$
Can you can break $mathbb{D}$ into two open charts and conclude that if it's constant on those charts and where the charts overlap then it must be constant on the entire set? I want to puncture the circle at a single point, conclude it's constant on that open set, then do it again with a different point and then conclude it's constant everywhere.
$endgroup$
– CyclotomicField
Nov 28 '18 at 18:30












$begingroup$
The other answers are slick (and correct), but I think I have a from scratch proof that might be what you are looking for.
$endgroup$
– Matematleta
Nov 28 '18 at 22:33




$begingroup$
The other answers are slick (and correct), but I think I have a from scratch proof that might be what you are looking for.
$endgroup$
– Matematleta
Nov 28 '18 at 22:33










4 Answers
4






active

oldest

votes


















3












$begingroup$

Let $B(0,1)$ be the unit open ball centered at the origin, and let $D^2 = B(0,1) cup mathbb{D}$ be the closed unit ball centered at the origin.



If $f$ is not constant, then by the open mapping theorem, $f(B(0,1)) = Omega$ is an open subset of $mathbb{C}$. Since $mathbb{D}$ is the boundary of $B(0,1)$ it maps into the boundary of $Omega$. Since $D^2$ is compact, $f(D^2)$ is compact; so $f(D^2)$ is closed and bounded.



So, $mathbb{D}$ maps onto the boundary of $Omega$. But this can only happen if $f(B(0,1)) = mathbb{C} - f(mathbb{D})$. This is indeed open because $f(mathbb{D})$ is a compact set and hence closed and bounded. However, this means that $f(B(0,1))$ is not bounded, which is a contradiction because $f(D^2)$ was deduced to be bounded, and $f(D^2) supset f(B(0,1))$.



Thus, we have a contradiction. Hence, $f$ must be constant.





By the way, your notation is a bit odd because I have seen $mathbb{D}$ to be usually used to denote the closed unit disc, rather than the circle. Perhaps it would be appropriate to double check the source of the question.






share|cite|improve this answer











$endgroup$





















    2












    $begingroup$

    If we set
    $$
    u(x,y)=mathrm{Im},f(x+iy),
    $$

    then $u$ is a harmonic function, which vanishes on the unit circle. Maximum (together with Minimum) Principle for harmonic function imply that
    $$
    max_{x^2+y^2le 1}{u(x,y)}=max_{x^2+y^2=1}{u(x,y)}=0
    quadtext{and}quad
    min_{x^2+y^2le 1}{u(x,y)}=min_{x^2+y^2=1}{u(x,y)}=0
    $$

    Hence, $uequiv 0$ in the closed unit disk, and hence $uequiv 0$ everywhere. (Identity Principle.)



    Thus $f$ takes only real values, and hence $f$ is constant.






    share|cite|improve this answer









    $endgroup$





















      2












      $begingroup$

      I think we can do this from scratch, especially since we are given that $f$ is a polynomial:



      Let $mathbb T$ be the unit circle, and $f$ a polynomial of degree $n$.



      For $z=e^{it}$, consider the real number



      $f(e^{it})=a_0 +a_1e^{it}+a_2e^{2it}+cdots a_{n}e^{nit}.$



      The imaginary part of this must be equal to zero, so



      $Im a_0+sum^{n}_{k=1}(Re a_ksin kt+Im a_kcos kt)=0$.



      But now, linear independence of the set $left {1, sin kt,cos kt right }^{n}_{k=1}$ on $mathbb T$ implies that



      $Im a_0=a_1=cdots=a_n=0$ so $f(z)=Re a_0$.






      share|cite|improve this answer











      $endgroup$





















        2












        $begingroup$

        The imaginary part $text{Im};f(z)$ of $f(z)$ is harmonic, and zero on the unit circle, hence by the maximum principle for harmonic functions, we conclude that $text{Im};f(z)$ is identically zero in the ball. Thus $f$ is real in the ball. Of course a non-constant real function cannot have complex derivative at $0$; in particular $f$ is not a polynomial.






        share|cite|improve this answer











        $endgroup$













        • $begingroup$
          Nice proof but I think it defeats the purpose of the questionbecause I doubt the OP has gotten that far
          $endgroup$
          – Matematleta
          Nov 28 '18 at 22:28






        • 1




          $begingroup$
          If he has not gotten to the maximum principle, I doubt he haa gotten to the open mapping theorem, either.
          $endgroup$
          – GEdgar
          Nov 28 '18 at 22:33










        • $begingroup$
          Fair enough, but since $f$ is given as a polynomial, it just seems like overkill to bring out the open mapping theorem/max modulus principle. I think you only need linear independence of ${1,e^{int}}$ on $mathbb T.$ I am no expert, though, so if you really do need these bigger guns, please let me know!
          $endgroup$
          – Matematleta
          Nov 28 '18 at 22:37












        • $begingroup$
          Nice proof using the maximum principle for harminic functions.
          $endgroup$
          – 674123173797 - 4
          Dec 1 '18 at 22:02











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        4 Answers
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        4 Answers
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        active

        oldest

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        active

        oldest

        votes






        active

        oldest

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        3












        $begingroup$

        Let $B(0,1)$ be the unit open ball centered at the origin, and let $D^2 = B(0,1) cup mathbb{D}$ be the closed unit ball centered at the origin.



        If $f$ is not constant, then by the open mapping theorem, $f(B(0,1)) = Omega$ is an open subset of $mathbb{C}$. Since $mathbb{D}$ is the boundary of $B(0,1)$ it maps into the boundary of $Omega$. Since $D^2$ is compact, $f(D^2)$ is compact; so $f(D^2)$ is closed and bounded.



        So, $mathbb{D}$ maps onto the boundary of $Omega$. But this can only happen if $f(B(0,1)) = mathbb{C} - f(mathbb{D})$. This is indeed open because $f(mathbb{D})$ is a compact set and hence closed and bounded. However, this means that $f(B(0,1))$ is not bounded, which is a contradiction because $f(D^2)$ was deduced to be bounded, and $f(D^2) supset f(B(0,1))$.



        Thus, we have a contradiction. Hence, $f$ must be constant.





        By the way, your notation is a bit odd because I have seen $mathbb{D}$ to be usually used to denote the closed unit disc, rather than the circle. Perhaps it would be appropriate to double check the source of the question.






        share|cite|improve this answer











        $endgroup$


















          3












          $begingroup$

          Let $B(0,1)$ be the unit open ball centered at the origin, and let $D^2 = B(0,1) cup mathbb{D}$ be the closed unit ball centered at the origin.



          If $f$ is not constant, then by the open mapping theorem, $f(B(0,1)) = Omega$ is an open subset of $mathbb{C}$. Since $mathbb{D}$ is the boundary of $B(0,1)$ it maps into the boundary of $Omega$. Since $D^2$ is compact, $f(D^2)$ is compact; so $f(D^2)$ is closed and bounded.



          So, $mathbb{D}$ maps onto the boundary of $Omega$. But this can only happen if $f(B(0,1)) = mathbb{C} - f(mathbb{D})$. This is indeed open because $f(mathbb{D})$ is a compact set and hence closed and bounded. However, this means that $f(B(0,1))$ is not bounded, which is a contradiction because $f(D^2)$ was deduced to be bounded, and $f(D^2) supset f(B(0,1))$.



          Thus, we have a contradiction. Hence, $f$ must be constant.





          By the way, your notation is a bit odd because I have seen $mathbb{D}$ to be usually used to denote the closed unit disc, rather than the circle. Perhaps it would be appropriate to double check the source of the question.






          share|cite|improve this answer











          $endgroup$
















            3












            3








            3





            $begingroup$

            Let $B(0,1)$ be the unit open ball centered at the origin, and let $D^2 = B(0,1) cup mathbb{D}$ be the closed unit ball centered at the origin.



            If $f$ is not constant, then by the open mapping theorem, $f(B(0,1)) = Omega$ is an open subset of $mathbb{C}$. Since $mathbb{D}$ is the boundary of $B(0,1)$ it maps into the boundary of $Omega$. Since $D^2$ is compact, $f(D^2)$ is compact; so $f(D^2)$ is closed and bounded.



            So, $mathbb{D}$ maps onto the boundary of $Omega$. But this can only happen if $f(B(0,1)) = mathbb{C} - f(mathbb{D})$. This is indeed open because $f(mathbb{D})$ is a compact set and hence closed and bounded. However, this means that $f(B(0,1))$ is not bounded, which is a contradiction because $f(D^2)$ was deduced to be bounded, and $f(D^2) supset f(B(0,1))$.



            Thus, we have a contradiction. Hence, $f$ must be constant.





            By the way, your notation is a bit odd because I have seen $mathbb{D}$ to be usually used to denote the closed unit disc, rather than the circle. Perhaps it would be appropriate to double check the source of the question.






            share|cite|improve this answer











            $endgroup$



            Let $B(0,1)$ be the unit open ball centered at the origin, and let $D^2 = B(0,1) cup mathbb{D}$ be the closed unit ball centered at the origin.



            If $f$ is not constant, then by the open mapping theorem, $f(B(0,1)) = Omega$ is an open subset of $mathbb{C}$. Since $mathbb{D}$ is the boundary of $B(0,1)$ it maps into the boundary of $Omega$. Since $D^2$ is compact, $f(D^2)$ is compact; so $f(D^2)$ is closed and bounded.



            So, $mathbb{D}$ maps onto the boundary of $Omega$. But this can only happen if $f(B(0,1)) = mathbb{C} - f(mathbb{D})$. This is indeed open because $f(mathbb{D})$ is a compact set and hence closed and bounded. However, this means that $f(B(0,1))$ is not bounded, which is a contradiction because $f(D^2)$ was deduced to be bounded, and $f(D^2) supset f(B(0,1))$.



            Thus, we have a contradiction. Hence, $f$ must be constant.





            By the way, your notation is a bit odd because I have seen $mathbb{D}$ to be usually used to denote the closed unit disc, rather than the circle. Perhaps it would be appropriate to double check the source of the question.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 28 '18 at 18:31

























            answered Nov 28 '18 at 18:21









            BrahadeeshBrahadeesh

            6,18742361




            6,18742361























                2












                $begingroup$

                If we set
                $$
                u(x,y)=mathrm{Im},f(x+iy),
                $$

                then $u$ is a harmonic function, which vanishes on the unit circle. Maximum (together with Minimum) Principle for harmonic function imply that
                $$
                max_{x^2+y^2le 1}{u(x,y)}=max_{x^2+y^2=1}{u(x,y)}=0
                quadtext{and}quad
                min_{x^2+y^2le 1}{u(x,y)}=min_{x^2+y^2=1}{u(x,y)}=0
                $$

                Hence, $uequiv 0$ in the closed unit disk, and hence $uequiv 0$ everywhere. (Identity Principle.)



                Thus $f$ takes only real values, and hence $f$ is constant.






                share|cite|improve this answer









                $endgroup$


















                  2












                  $begingroup$

                  If we set
                  $$
                  u(x,y)=mathrm{Im},f(x+iy),
                  $$

                  then $u$ is a harmonic function, which vanishes on the unit circle. Maximum (together with Minimum) Principle for harmonic function imply that
                  $$
                  max_{x^2+y^2le 1}{u(x,y)}=max_{x^2+y^2=1}{u(x,y)}=0
                  quadtext{and}quad
                  min_{x^2+y^2le 1}{u(x,y)}=min_{x^2+y^2=1}{u(x,y)}=0
                  $$

                  Hence, $uequiv 0$ in the closed unit disk, and hence $uequiv 0$ everywhere. (Identity Principle.)



                  Thus $f$ takes only real values, and hence $f$ is constant.






                  share|cite|improve this answer









                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    If we set
                    $$
                    u(x,y)=mathrm{Im},f(x+iy),
                    $$

                    then $u$ is a harmonic function, which vanishes on the unit circle. Maximum (together with Minimum) Principle for harmonic function imply that
                    $$
                    max_{x^2+y^2le 1}{u(x,y)}=max_{x^2+y^2=1}{u(x,y)}=0
                    quadtext{and}quad
                    min_{x^2+y^2le 1}{u(x,y)}=min_{x^2+y^2=1}{u(x,y)}=0
                    $$

                    Hence, $uequiv 0$ in the closed unit disk, and hence $uequiv 0$ everywhere. (Identity Principle.)



                    Thus $f$ takes only real values, and hence $f$ is constant.






                    share|cite|improve this answer









                    $endgroup$



                    If we set
                    $$
                    u(x,y)=mathrm{Im},f(x+iy),
                    $$

                    then $u$ is a harmonic function, which vanishes on the unit circle. Maximum (together with Minimum) Principle for harmonic function imply that
                    $$
                    max_{x^2+y^2le 1}{u(x,y)}=max_{x^2+y^2=1}{u(x,y)}=0
                    quadtext{and}quad
                    min_{x^2+y^2le 1}{u(x,y)}=min_{x^2+y^2=1}{u(x,y)}=0
                    $$

                    Hence, $uequiv 0$ in the closed unit disk, and hence $uequiv 0$ everywhere. (Identity Principle.)



                    Thus $f$ takes only real values, and hence $f$ is constant.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Nov 28 '18 at 22:25









                    Yiorgos S. SmyrlisYiorgos S. Smyrlis

                    62.9k1384163




                    62.9k1384163























                        2












                        $begingroup$

                        I think we can do this from scratch, especially since we are given that $f$ is a polynomial:



                        Let $mathbb T$ be the unit circle, and $f$ a polynomial of degree $n$.



                        For $z=e^{it}$, consider the real number



                        $f(e^{it})=a_0 +a_1e^{it}+a_2e^{2it}+cdots a_{n}e^{nit}.$



                        The imaginary part of this must be equal to zero, so



                        $Im a_0+sum^{n}_{k=1}(Re a_ksin kt+Im a_kcos kt)=0$.



                        But now, linear independence of the set $left {1, sin kt,cos kt right }^{n}_{k=1}$ on $mathbb T$ implies that



                        $Im a_0=a_1=cdots=a_n=0$ so $f(z)=Re a_0$.






                        share|cite|improve this answer











                        $endgroup$


















                          2












                          $begingroup$

                          I think we can do this from scratch, especially since we are given that $f$ is a polynomial:



                          Let $mathbb T$ be the unit circle, and $f$ a polynomial of degree $n$.



                          For $z=e^{it}$, consider the real number



                          $f(e^{it})=a_0 +a_1e^{it}+a_2e^{2it}+cdots a_{n}e^{nit}.$



                          The imaginary part of this must be equal to zero, so



                          $Im a_0+sum^{n}_{k=1}(Re a_ksin kt+Im a_kcos kt)=0$.



                          But now, linear independence of the set $left {1, sin kt,cos kt right }^{n}_{k=1}$ on $mathbb T$ implies that



                          $Im a_0=a_1=cdots=a_n=0$ so $f(z)=Re a_0$.






                          share|cite|improve this answer











                          $endgroup$
















                            2












                            2








                            2





                            $begingroup$

                            I think we can do this from scratch, especially since we are given that $f$ is a polynomial:



                            Let $mathbb T$ be the unit circle, and $f$ a polynomial of degree $n$.



                            For $z=e^{it}$, consider the real number



                            $f(e^{it})=a_0 +a_1e^{it}+a_2e^{2it}+cdots a_{n}e^{nit}.$



                            The imaginary part of this must be equal to zero, so



                            $Im a_0+sum^{n}_{k=1}(Re a_ksin kt+Im a_kcos kt)=0$.



                            But now, linear independence of the set $left {1, sin kt,cos kt right }^{n}_{k=1}$ on $mathbb T$ implies that



                            $Im a_0=a_1=cdots=a_n=0$ so $f(z)=Re a_0$.






                            share|cite|improve this answer











                            $endgroup$



                            I think we can do this from scratch, especially since we are given that $f$ is a polynomial:



                            Let $mathbb T$ be the unit circle, and $f$ a polynomial of degree $n$.



                            For $z=e^{it}$, consider the real number



                            $f(e^{it})=a_0 +a_1e^{it}+a_2e^{2it}+cdots a_{n}e^{nit}.$



                            The imaginary part of this must be equal to zero, so



                            $Im a_0+sum^{n}_{k=1}(Re a_ksin kt+Im a_kcos kt)=0$.



                            But now, linear independence of the set $left {1, sin kt,cos kt right }^{n}_{k=1}$ on $mathbb T$ implies that



                            $Im a_0=a_1=cdots=a_n=0$ so $f(z)=Re a_0$.







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Nov 28 '18 at 22:54

























                            answered Nov 28 '18 at 17:16









                            MatematletaMatematleta

                            10.2k2918




                            10.2k2918























                                2












                                $begingroup$

                                The imaginary part $text{Im};f(z)$ of $f(z)$ is harmonic, and zero on the unit circle, hence by the maximum principle for harmonic functions, we conclude that $text{Im};f(z)$ is identically zero in the ball. Thus $f$ is real in the ball. Of course a non-constant real function cannot have complex derivative at $0$; in particular $f$ is not a polynomial.






                                share|cite|improve this answer











                                $endgroup$













                                • $begingroup$
                                  Nice proof but I think it defeats the purpose of the questionbecause I doubt the OP has gotten that far
                                  $endgroup$
                                  – Matematleta
                                  Nov 28 '18 at 22:28






                                • 1




                                  $begingroup$
                                  If he has not gotten to the maximum principle, I doubt he haa gotten to the open mapping theorem, either.
                                  $endgroup$
                                  – GEdgar
                                  Nov 28 '18 at 22:33










                                • $begingroup$
                                  Fair enough, but since $f$ is given as a polynomial, it just seems like overkill to bring out the open mapping theorem/max modulus principle. I think you only need linear independence of ${1,e^{int}}$ on $mathbb T.$ I am no expert, though, so if you really do need these bigger guns, please let me know!
                                  $endgroup$
                                  – Matematleta
                                  Nov 28 '18 at 22:37












                                • $begingroup$
                                  Nice proof using the maximum principle for harminic functions.
                                  $endgroup$
                                  – 674123173797 - 4
                                  Dec 1 '18 at 22:02
















                                2












                                $begingroup$

                                The imaginary part $text{Im};f(z)$ of $f(z)$ is harmonic, and zero on the unit circle, hence by the maximum principle for harmonic functions, we conclude that $text{Im};f(z)$ is identically zero in the ball. Thus $f$ is real in the ball. Of course a non-constant real function cannot have complex derivative at $0$; in particular $f$ is not a polynomial.






                                share|cite|improve this answer











                                $endgroup$













                                • $begingroup$
                                  Nice proof but I think it defeats the purpose of the questionbecause I doubt the OP has gotten that far
                                  $endgroup$
                                  – Matematleta
                                  Nov 28 '18 at 22:28






                                • 1




                                  $begingroup$
                                  If he has not gotten to the maximum principle, I doubt he haa gotten to the open mapping theorem, either.
                                  $endgroup$
                                  – GEdgar
                                  Nov 28 '18 at 22:33










                                • $begingroup$
                                  Fair enough, but since $f$ is given as a polynomial, it just seems like overkill to bring out the open mapping theorem/max modulus principle. I think you only need linear independence of ${1,e^{int}}$ on $mathbb T.$ I am no expert, though, so if you really do need these bigger guns, please let me know!
                                  $endgroup$
                                  – Matematleta
                                  Nov 28 '18 at 22:37












                                • $begingroup$
                                  Nice proof using the maximum principle for harminic functions.
                                  $endgroup$
                                  – 674123173797 - 4
                                  Dec 1 '18 at 22:02














                                2












                                2








                                2





                                $begingroup$

                                The imaginary part $text{Im};f(z)$ of $f(z)$ is harmonic, and zero on the unit circle, hence by the maximum principle for harmonic functions, we conclude that $text{Im};f(z)$ is identically zero in the ball. Thus $f$ is real in the ball. Of course a non-constant real function cannot have complex derivative at $0$; in particular $f$ is not a polynomial.






                                share|cite|improve this answer











                                $endgroup$



                                The imaginary part $text{Im};f(z)$ of $f(z)$ is harmonic, and zero on the unit circle, hence by the maximum principle for harmonic functions, we conclude that $text{Im};f(z)$ is identically zero in the ball. Thus $f$ is real in the ball. Of course a non-constant real function cannot have complex derivative at $0$; in particular $f$ is not a polynomial.







                                share|cite|improve this answer














                                share|cite|improve this answer



                                share|cite|improve this answer








                                edited Dec 1 '18 at 22:36

























                                answered Nov 28 '18 at 22:26









                                GEdgarGEdgar

                                62k267168




                                62k267168












                                • $begingroup$
                                  Nice proof but I think it defeats the purpose of the questionbecause I doubt the OP has gotten that far
                                  $endgroup$
                                  – Matematleta
                                  Nov 28 '18 at 22:28






                                • 1




                                  $begingroup$
                                  If he has not gotten to the maximum principle, I doubt he haa gotten to the open mapping theorem, either.
                                  $endgroup$
                                  – GEdgar
                                  Nov 28 '18 at 22:33










                                • $begingroup$
                                  Fair enough, but since $f$ is given as a polynomial, it just seems like overkill to bring out the open mapping theorem/max modulus principle. I think you only need linear independence of ${1,e^{int}}$ on $mathbb T.$ I am no expert, though, so if you really do need these bigger guns, please let me know!
                                  $endgroup$
                                  – Matematleta
                                  Nov 28 '18 at 22:37












                                • $begingroup$
                                  Nice proof using the maximum principle for harminic functions.
                                  $endgroup$
                                  – 674123173797 - 4
                                  Dec 1 '18 at 22:02


















                                • $begingroup$
                                  Nice proof but I think it defeats the purpose of the questionbecause I doubt the OP has gotten that far
                                  $endgroup$
                                  – Matematleta
                                  Nov 28 '18 at 22:28






                                • 1




                                  $begingroup$
                                  If he has not gotten to the maximum principle, I doubt he haa gotten to the open mapping theorem, either.
                                  $endgroup$
                                  – GEdgar
                                  Nov 28 '18 at 22:33










                                • $begingroup$
                                  Fair enough, but since $f$ is given as a polynomial, it just seems like overkill to bring out the open mapping theorem/max modulus principle. I think you only need linear independence of ${1,e^{int}}$ on $mathbb T.$ I am no expert, though, so if you really do need these bigger guns, please let me know!
                                  $endgroup$
                                  – Matematleta
                                  Nov 28 '18 at 22:37












                                • $begingroup$
                                  Nice proof using the maximum principle for harminic functions.
                                  $endgroup$
                                  – 674123173797 - 4
                                  Dec 1 '18 at 22:02
















                                $begingroup$
                                Nice proof but I think it defeats the purpose of the questionbecause I doubt the OP has gotten that far
                                $endgroup$
                                – Matematleta
                                Nov 28 '18 at 22:28




                                $begingroup$
                                Nice proof but I think it defeats the purpose of the questionbecause I doubt the OP has gotten that far
                                $endgroup$
                                – Matematleta
                                Nov 28 '18 at 22:28




                                1




                                1




                                $begingroup$
                                If he has not gotten to the maximum principle, I doubt he haa gotten to the open mapping theorem, either.
                                $endgroup$
                                – GEdgar
                                Nov 28 '18 at 22:33




                                $begingroup$
                                If he has not gotten to the maximum principle, I doubt he haa gotten to the open mapping theorem, either.
                                $endgroup$
                                – GEdgar
                                Nov 28 '18 at 22:33












                                $begingroup$
                                Fair enough, but since $f$ is given as a polynomial, it just seems like overkill to bring out the open mapping theorem/max modulus principle. I think you only need linear independence of ${1,e^{int}}$ on $mathbb T.$ I am no expert, though, so if you really do need these bigger guns, please let me know!
                                $endgroup$
                                – Matematleta
                                Nov 28 '18 at 22:37






                                $begingroup$
                                Fair enough, but since $f$ is given as a polynomial, it just seems like overkill to bring out the open mapping theorem/max modulus principle. I think you only need linear independence of ${1,e^{int}}$ on $mathbb T.$ I am no expert, though, so if you really do need these bigger guns, please let me know!
                                $endgroup$
                                – Matematleta
                                Nov 28 '18 at 22:37














                                $begingroup$
                                Nice proof using the maximum principle for harminic functions.
                                $endgroup$
                                – 674123173797 - 4
                                Dec 1 '18 at 22:02




                                $begingroup$
                                Nice proof using the maximum principle for harminic functions.
                                $endgroup$
                                – 674123173797 - 4
                                Dec 1 '18 at 22:02


















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