Does there exist an analytic function $f$ on a domain containing the unit disk such that $f(z) = exp(i,...












2












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Does there exist an analytic function $f:Drightarrowmathbb C$, where $D$ is a domain containing the unit disk, such that $f(z) = e^{i text{Im} z}$ on the unit circle $|z|=1$?




I'm supposed to rely on the fact that if $f: B_R=left { |z|leq R right }rightarrow mathbb C$ is analytic and even on $(-R,R)$ (so $f(x)=f(-x)$ when $x$ is real), then it is also even on $B_R$, outside the real number line.



It seems that we need to show that $f$ isn't even on $B_1$ but is even on $(-1,1)$, which I cannot show. We only know how $f$ behaves on the unit circle!






complex-analysis







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  • 1




    $begingroup$
    Please avoid cryptic titles.
    $endgroup$
    – Did
    Dec 30 '18 at 21:08






  • 1




    $begingroup$
    It's better not to use $R$ to mean two different things.
    $endgroup$
    – Greg Martin
    Dec 30 '18 at 21:26
















2












$begingroup$



Does there exist an analytic function $f:Drightarrowmathbb C$, where $D$ is a domain containing the unit disk, such that $f(z) = e^{i text{Im} z}$ on the unit circle $|z|=1$?




I'm supposed to rely on the fact that if $f: B_R=left { |z|leq R right }rightarrow mathbb C$ is analytic and even on $(-R,R)$ (so $f(x)=f(-x)$ when $x$ is real), then it is also even on $B_R$, outside the real number line.



It seems that we need to show that $f$ isn't even on $B_1$ but is even on $(-1,1)$, which I cannot show. We only know how $f$ behaves on the unit circle!






complex-analysis







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Please avoid cryptic titles.
    $endgroup$
    – Did
    Dec 30 '18 at 21:08






  • 1




    $begingroup$
    It's better not to use $R$ to mean two different things.
    $endgroup$
    – Greg Martin
    Dec 30 '18 at 21:26














2












2








2





$begingroup$



Does there exist an analytic function $f:Drightarrowmathbb C$, where $D$ is a domain containing the unit disk, such that $f(z) = e^{i text{Im} z}$ on the unit circle $|z|=1$?




I'm supposed to rely on the fact that if $f: B_R=left { |z|leq R right }rightarrow mathbb C$ is analytic and even on $(-R,R)$ (so $f(x)=f(-x)$ when $x$ is real), then it is also even on $B_R$, outside the real number line.



It seems that we need to show that $f$ isn't even on $B_1$ but is even on $(-1,1)$, which I cannot show. We only know how $f$ behaves on the unit circle!






complex-analysis







share|cite|improve this question











$endgroup$





Does there exist an analytic function $f:Drightarrowmathbb C$, where $D$ is a domain containing the unit disk, such that $f(z) = e^{i text{Im} z}$ on the unit circle $|z|=1$?




I'm supposed to rely on the fact that if $f: B_R=left { |z|leq R right }rightarrow mathbb C$ is analytic and even on $(-R,R)$ (so $f(x)=f(-x)$ when $x$ is real), then it is also even on $B_R$, outside the real number line.



It seems that we need to show that $f$ isn't even on $B_1$ but is even on $(-1,1)$, which I cannot show. We only know how $f$ behaves on the unit circle!






complex-analysis




complex-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 30 '18 at 21:38







Theorem

















asked Dec 30 '18 at 21:04









TheoremTheorem

1,061513




1,061513








  • 1




    $begingroup$
    Please avoid cryptic titles.
    $endgroup$
    – Did
    Dec 30 '18 at 21:08






  • 1




    $begingroup$
    It's better not to use $R$ to mean two different things.
    $endgroup$
    – Greg Martin
    Dec 30 '18 at 21:26














  • 1




    $begingroup$
    Please avoid cryptic titles.
    $endgroup$
    – Did
    Dec 30 '18 at 21:08






  • 1




    $begingroup$
    It's better not to use $R$ to mean two different things.
    $endgroup$
    – Greg Martin
    Dec 30 '18 at 21:26








1




1




$begingroup$
Please avoid cryptic titles.
$endgroup$
– Did
Dec 30 '18 at 21:08




$begingroup$
Please avoid cryptic titles.
$endgroup$
– Did
Dec 30 '18 at 21:08




1




1




$begingroup$
It's better not to use $R$ to mean two different things.
$endgroup$
– Greg Martin
Dec 30 '18 at 21:26




$begingroup$
It's better not to use $R$ to mean two different things.
$endgroup$
– Greg Martin
Dec 30 '18 at 21:26










3 Answers
3






active

oldest

votes


















3












$begingroup$

Here is an answer which only uses your hint in spirit. Ideally you should think about Greg Martin's hints before reading this, since they are of a similar spirit.



You know that $f(z)$ satisfies the relation $f(z)f(1/z) = 1$ on the unit circle; because $g(z) = f(z)f(1/z) -1$ is a holomorphic function defined in a neighborhood of the circle, and it vanishes on a set with an accumulation point, it vanishes everywhere on that neighborhood. By the same logic $f(z) f(-z) = 1$ on the unit disc as well.



In particular, the second formula implies $f(z)$ is nowhere everywhere along the unit disc. This allows us to use the first formula to give $f$ an extension to the entire complex plane; for $|z| > 1$, set $f(z) = 1/f(1/z)$. Because the RHS is already holomorphic, and this is true in the neighborhood of the unit circle, these patch together to gives us a formula for a globally defined extension of $f$ (which I will denote by the same name).



Because $1/f$ is continuous and nonzero on the unit disc, we see that $1/f$ achieves a maximum, and hence that $f(1/z)$ achieves a maximum on $Bbb C setminus Bbb D$.



Combining these two maximums, we see that $|f|$ is bounded above. Thus $f$ is a bounded holomorphic function, and hence constant, contradicting your formula on the circle.






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  • $begingroup$
    Wait.. how do we know $f(z)f(1/z)-1$ is holomorphic? We only know it vanishes on the unit circle
    $endgroup$
    – Theorem
    Dec 30 '18 at 22:03










  • $begingroup$
    @Theorem a composition of holomorphic functions is holomorphic, a product of holomorphic functions is holomorphic, and a sum of holomorphic with functions is holomorphic. What I should really say is $g$ has domain $text{Dom}(f) setminus {0}$, but this is not really a big deal; the main use of this formula is to show that $f$ is naturally defined on all of $Bbb C$, and for that application you merely need $g$ to be holomorphic in a neighborhood of that circle.
    $endgroup$
    – user98602
    Dec 30 '18 at 22:04












  • $begingroup$
    This is a nice answer under the additional assumption that $f$ is entire; however, the OP does not assume that. Your proof uses that assumption twice, once when considering the domain of $g(z)$ via $f(1/z)$ and again when invoking facts about bounded functions.
    $endgroup$
    – Greg Martin
    Dec 31 '18 at 7:33










  • $begingroup$
    @GregMartin It does not use this assumption but I think I didn't explain that clearly enough. One sees that on a neighborhood of $S^1$ contained in the domain of $f$, that $g$ is holomorphic in that neighood and vanishes along the circle; so $f(1/z) = 1/f(z)$ in that nbhd. One may naively extend $f$ to the entire complex plane by the rule $f(1/z) = 1/f(z)$ for $|z| > 1$. Because this formula is true near the circle (where $f$ is holomorphic), and both the given def'ns for $f$ on the int. and ext. of the circle are holomorphic, we see that $f$ has a canonical extension to an entire function.
    $endgroup$
    – user98602
    Dec 31 '18 at 7:38










  • $begingroup$
    (It was precisely in obtaining the entire extension that we used that $f$ was holomorphic on the entire disc.) As a side comment, I originally wanted to use the Schwarz reflection principle, since $f(S^1) subset S^1$; that also shows that $f$ has a holomorphic extension to all of $Bbb C$. But then I saw there was a more elementary way to see that in this case.
    $endgroup$
    – user98602
    Dec 31 '18 at 7:45





















3












$begingroup$

Sequence of hints:




  • If $f$ and $g$ are two such functions, show they must be equal.

  • If $f(z)$ is such a function, show that $g(z) = overline{f(bar z)}$ (its "reflection in the real axis") is another such function.

  • If $f(z)$ is such a function, show that $f(x)$ is real-valued for all $xin(-1,1)$.

  • If $f(z)$ is such a function, show that $h(z) = overline{f(-bar z)}$ (its "reflection in the imaginary axis") is another such function.

  • If $f(z)$ is such a function, show that $f(x)=f(-x)$ for all $xin(-1,1)$.






share|cite|improve this answer









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    3












    $begingroup$

    Nope. $$oint_{|z|=1} e^{iIm z} dz = iint_0^{2pi} e^{i(sintheta + theta)} dtheta = -2pi iJ_1(1) ne 0$$






    share|cite|improve this answer









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      3 Answers
      3






      active

      oldest

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      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      Here is an answer which only uses your hint in spirit. Ideally you should think about Greg Martin's hints before reading this, since they are of a similar spirit.



      You know that $f(z)$ satisfies the relation $f(z)f(1/z) = 1$ on the unit circle; because $g(z) = f(z)f(1/z) -1$ is a holomorphic function defined in a neighborhood of the circle, and it vanishes on a set with an accumulation point, it vanishes everywhere on that neighborhood. By the same logic $f(z) f(-z) = 1$ on the unit disc as well.



      In particular, the second formula implies $f(z)$ is nowhere everywhere along the unit disc. This allows us to use the first formula to give $f$ an extension to the entire complex plane; for $|z| > 1$, set $f(z) = 1/f(1/z)$. Because the RHS is already holomorphic, and this is true in the neighborhood of the unit circle, these patch together to gives us a formula for a globally defined extension of $f$ (which I will denote by the same name).



      Because $1/f$ is continuous and nonzero on the unit disc, we see that $1/f$ achieves a maximum, and hence that $f(1/z)$ achieves a maximum on $Bbb C setminus Bbb D$.



      Combining these two maximums, we see that $|f|$ is bounded above. Thus $f$ is a bounded holomorphic function, and hence constant, contradicting your formula on the circle.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        Wait.. how do we know $f(z)f(1/z)-1$ is holomorphic? We only know it vanishes on the unit circle
        $endgroup$
        – Theorem
        Dec 30 '18 at 22:03










      • $begingroup$
        @Theorem a composition of holomorphic functions is holomorphic, a product of holomorphic functions is holomorphic, and a sum of holomorphic with functions is holomorphic. What I should really say is $g$ has domain $text{Dom}(f) setminus {0}$, but this is not really a big deal; the main use of this formula is to show that $f$ is naturally defined on all of $Bbb C$, and for that application you merely need $g$ to be holomorphic in a neighborhood of that circle.
        $endgroup$
        – user98602
        Dec 30 '18 at 22:04












      • $begingroup$
        This is a nice answer under the additional assumption that $f$ is entire; however, the OP does not assume that. Your proof uses that assumption twice, once when considering the domain of $g(z)$ via $f(1/z)$ and again when invoking facts about bounded functions.
        $endgroup$
        – Greg Martin
        Dec 31 '18 at 7:33










      • $begingroup$
        @GregMartin It does not use this assumption but I think I didn't explain that clearly enough. One sees that on a neighborhood of $S^1$ contained in the domain of $f$, that $g$ is holomorphic in that neighood and vanishes along the circle; so $f(1/z) = 1/f(z)$ in that nbhd. One may naively extend $f$ to the entire complex plane by the rule $f(1/z) = 1/f(z)$ for $|z| > 1$. Because this formula is true near the circle (where $f$ is holomorphic), and both the given def'ns for $f$ on the int. and ext. of the circle are holomorphic, we see that $f$ has a canonical extension to an entire function.
        $endgroup$
        – user98602
        Dec 31 '18 at 7:38










      • $begingroup$
        (It was precisely in obtaining the entire extension that we used that $f$ was holomorphic on the entire disc.) As a side comment, I originally wanted to use the Schwarz reflection principle, since $f(S^1) subset S^1$; that also shows that $f$ has a holomorphic extension to all of $Bbb C$. But then I saw there was a more elementary way to see that in this case.
        $endgroup$
        – user98602
        Dec 31 '18 at 7:45


















      3












      $begingroup$

      Here is an answer which only uses your hint in spirit. Ideally you should think about Greg Martin's hints before reading this, since they are of a similar spirit.



      You know that $f(z)$ satisfies the relation $f(z)f(1/z) = 1$ on the unit circle; because $g(z) = f(z)f(1/z) -1$ is a holomorphic function defined in a neighborhood of the circle, and it vanishes on a set with an accumulation point, it vanishes everywhere on that neighborhood. By the same logic $f(z) f(-z) = 1$ on the unit disc as well.



      In particular, the second formula implies $f(z)$ is nowhere everywhere along the unit disc. This allows us to use the first formula to give $f$ an extension to the entire complex plane; for $|z| > 1$, set $f(z) = 1/f(1/z)$. Because the RHS is already holomorphic, and this is true in the neighborhood of the unit circle, these patch together to gives us a formula for a globally defined extension of $f$ (which I will denote by the same name).



      Because $1/f$ is continuous and nonzero on the unit disc, we see that $1/f$ achieves a maximum, and hence that $f(1/z)$ achieves a maximum on $Bbb C setminus Bbb D$.



      Combining these two maximums, we see that $|f|$ is bounded above. Thus $f$ is a bounded holomorphic function, and hence constant, contradicting your formula on the circle.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        Wait.. how do we know $f(z)f(1/z)-1$ is holomorphic? We only know it vanishes on the unit circle
        $endgroup$
        – Theorem
        Dec 30 '18 at 22:03










      • $begingroup$
        @Theorem a composition of holomorphic functions is holomorphic, a product of holomorphic functions is holomorphic, and a sum of holomorphic with functions is holomorphic. What I should really say is $g$ has domain $text{Dom}(f) setminus {0}$, but this is not really a big deal; the main use of this formula is to show that $f$ is naturally defined on all of $Bbb C$, and for that application you merely need $g$ to be holomorphic in a neighborhood of that circle.
        $endgroup$
        – user98602
        Dec 30 '18 at 22:04












      • $begingroup$
        This is a nice answer under the additional assumption that $f$ is entire; however, the OP does not assume that. Your proof uses that assumption twice, once when considering the domain of $g(z)$ via $f(1/z)$ and again when invoking facts about bounded functions.
        $endgroup$
        – Greg Martin
        Dec 31 '18 at 7:33










      • $begingroup$
        @GregMartin It does not use this assumption but I think I didn't explain that clearly enough. One sees that on a neighborhood of $S^1$ contained in the domain of $f$, that $g$ is holomorphic in that neighood and vanishes along the circle; so $f(1/z) = 1/f(z)$ in that nbhd. One may naively extend $f$ to the entire complex plane by the rule $f(1/z) = 1/f(z)$ for $|z| > 1$. Because this formula is true near the circle (where $f$ is holomorphic), and both the given def'ns for $f$ on the int. and ext. of the circle are holomorphic, we see that $f$ has a canonical extension to an entire function.
        $endgroup$
        – user98602
        Dec 31 '18 at 7:38










      • $begingroup$
        (It was precisely in obtaining the entire extension that we used that $f$ was holomorphic on the entire disc.) As a side comment, I originally wanted to use the Schwarz reflection principle, since $f(S^1) subset S^1$; that also shows that $f$ has a holomorphic extension to all of $Bbb C$. But then I saw there was a more elementary way to see that in this case.
        $endgroup$
        – user98602
        Dec 31 '18 at 7:45
















      3












      3








      3





      $begingroup$

      Here is an answer which only uses your hint in spirit. Ideally you should think about Greg Martin's hints before reading this, since they are of a similar spirit.



      You know that $f(z)$ satisfies the relation $f(z)f(1/z) = 1$ on the unit circle; because $g(z) = f(z)f(1/z) -1$ is a holomorphic function defined in a neighborhood of the circle, and it vanishes on a set with an accumulation point, it vanishes everywhere on that neighborhood. By the same logic $f(z) f(-z) = 1$ on the unit disc as well.



      In particular, the second formula implies $f(z)$ is nowhere everywhere along the unit disc. This allows us to use the first formula to give $f$ an extension to the entire complex plane; for $|z| > 1$, set $f(z) = 1/f(1/z)$. Because the RHS is already holomorphic, and this is true in the neighborhood of the unit circle, these patch together to gives us a formula for a globally defined extension of $f$ (which I will denote by the same name).



      Because $1/f$ is continuous and nonzero on the unit disc, we see that $1/f$ achieves a maximum, and hence that $f(1/z)$ achieves a maximum on $Bbb C setminus Bbb D$.



      Combining these two maximums, we see that $|f|$ is bounded above. Thus $f$ is a bounded holomorphic function, and hence constant, contradicting your formula on the circle.






      share|cite|improve this answer











      $endgroup$



      Here is an answer which only uses your hint in spirit. Ideally you should think about Greg Martin's hints before reading this, since they are of a similar spirit.



      You know that $f(z)$ satisfies the relation $f(z)f(1/z) = 1$ on the unit circle; because $g(z) = f(z)f(1/z) -1$ is a holomorphic function defined in a neighborhood of the circle, and it vanishes on a set with an accumulation point, it vanishes everywhere on that neighborhood. By the same logic $f(z) f(-z) = 1$ on the unit disc as well.



      In particular, the second formula implies $f(z)$ is nowhere everywhere along the unit disc. This allows us to use the first formula to give $f$ an extension to the entire complex plane; for $|z| > 1$, set $f(z) = 1/f(1/z)$. Because the RHS is already holomorphic, and this is true in the neighborhood of the unit circle, these patch together to gives us a formula for a globally defined extension of $f$ (which I will denote by the same name).



      Because $1/f$ is continuous and nonzero on the unit disc, we see that $1/f$ achieves a maximum, and hence that $f(1/z)$ achieves a maximum on $Bbb C setminus Bbb D$.



      Combining these two maximums, we see that $|f|$ is bounded above. Thus $f$ is a bounded holomorphic function, and hence constant, contradicting your formula on the circle.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Dec 31 '18 at 13:08

























      answered Dec 30 '18 at 21:46







      user98602



















      • $begingroup$
        Wait.. how do we know $f(z)f(1/z)-1$ is holomorphic? We only know it vanishes on the unit circle
        $endgroup$
        – Theorem
        Dec 30 '18 at 22:03










      • $begingroup$
        @Theorem a composition of holomorphic functions is holomorphic, a product of holomorphic functions is holomorphic, and a sum of holomorphic with functions is holomorphic. What I should really say is $g$ has domain $text{Dom}(f) setminus {0}$, but this is not really a big deal; the main use of this formula is to show that $f$ is naturally defined on all of $Bbb C$, and for that application you merely need $g$ to be holomorphic in a neighborhood of that circle.
        $endgroup$
        – user98602
        Dec 30 '18 at 22:04












      • $begingroup$
        This is a nice answer under the additional assumption that $f$ is entire; however, the OP does not assume that. Your proof uses that assumption twice, once when considering the domain of $g(z)$ via $f(1/z)$ and again when invoking facts about bounded functions.
        $endgroup$
        – Greg Martin
        Dec 31 '18 at 7:33










      • $begingroup$
        @GregMartin It does not use this assumption but I think I didn't explain that clearly enough. One sees that on a neighborhood of $S^1$ contained in the domain of $f$, that $g$ is holomorphic in that neighood and vanishes along the circle; so $f(1/z) = 1/f(z)$ in that nbhd. One may naively extend $f$ to the entire complex plane by the rule $f(1/z) = 1/f(z)$ for $|z| > 1$. Because this formula is true near the circle (where $f$ is holomorphic), and both the given def'ns for $f$ on the int. and ext. of the circle are holomorphic, we see that $f$ has a canonical extension to an entire function.
        $endgroup$
        – user98602
        Dec 31 '18 at 7:38










      • $begingroup$
        (It was precisely in obtaining the entire extension that we used that $f$ was holomorphic on the entire disc.) As a side comment, I originally wanted to use the Schwarz reflection principle, since $f(S^1) subset S^1$; that also shows that $f$ has a holomorphic extension to all of $Bbb C$. But then I saw there was a more elementary way to see that in this case.
        $endgroup$
        – user98602
        Dec 31 '18 at 7:45




















      • $begingroup$
        Wait.. how do we know $f(z)f(1/z)-1$ is holomorphic? We only know it vanishes on the unit circle
        $endgroup$
        – Theorem
        Dec 30 '18 at 22:03










      • $begingroup$
        @Theorem a composition of holomorphic functions is holomorphic, a product of holomorphic functions is holomorphic, and a sum of holomorphic with functions is holomorphic. What I should really say is $g$ has domain $text{Dom}(f) setminus {0}$, but this is not really a big deal; the main use of this formula is to show that $f$ is naturally defined on all of $Bbb C$, and for that application you merely need $g$ to be holomorphic in a neighborhood of that circle.
        $endgroup$
        – user98602
        Dec 30 '18 at 22:04












      • $begingroup$
        This is a nice answer under the additional assumption that $f$ is entire; however, the OP does not assume that. Your proof uses that assumption twice, once when considering the domain of $g(z)$ via $f(1/z)$ and again when invoking facts about bounded functions.
        $endgroup$
        – Greg Martin
        Dec 31 '18 at 7:33










      • $begingroup$
        @GregMartin It does not use this assumption but I think I didn't explain that clearly enough. One sees that on a neighborhood of $S^1$ contained in the domain of $f$, that $g$ is holomorphic in that neighood and vanishes along the circle; so $f(1/z) = 1/f(z)$ in that nbhd. One may naively extend $f$ to the entire complex plane by the rule $f(1/z) = 1/f(z)$ for $|z| > 1$. Because this formula is true near the circle (where $f$ is holomorphic), and both the given def'ns for $f$ on the int. and ext. of the circle are holomorphic, we see that $f$ has a canonical extension to an entire function.
        $endgroup$
        – user98602
        Dec 31 '18 at 7:38










      • $begingroup$
        (It was precisely in obtaining the entire extension that we used that $f$ was holomorphic on the entire disc.) As a side comment, I originally wanted to use the Schwarz reflection principle, since $f(S^1) subset S^1$; that also shows that $f$ has a holomorphic extension to all of $Bbb C$. But then I saw there was a more elementary way to see that in this case.
        $endgroup$
        – user98602
        Dec 31 '18 at 7:45


















      $begingroup$
      Wait.. how do we know $f(z)f(1/z)-1$ is holomorphic? We only know it vanishes on the unit circle
      $endgroup$
      – Theorem
      Dec 30 '18 at 22:03




      $begingroup$
      Wait.. how do we know $f(z)f(1/z)-1$ is holomorphic? We only know it vanishes on the unit circle
      $endgroup$
      – Theorem
      Dec 30 '18 at 22:03












      $begingroup$
      @Theorem a composition of holomorphic functions is holomorphic, a product of holomorphic functions is holomorphic, and a sum of holomorphic with functions is holomorphic. What I should really say is $g$ has domain $text{Dom}(f) setminus {0}$, but this is not really a big deal; the main use of this formula is to show that $f$ is naturally defined on all of $Bbb C$, and for that application you merely need $g$ to be holomorphic in a neighborhood of that circle.
      $endgroup$
      – user98602
      Dec 30 '18 at 22:04






      $begingroup$
      @Theorem a composition of holomorphic functions is holomorphic, a product of holomorphic functions is holomorphic, and a sum of holomorphic with functions is holomorphic. What I should really say is $g$ has domain $text{Dom}(f) setminus {0}$, but this is not really a big deal; the main use of this formula is to show that $f$ is naturally defined on all of $Bbb C$, and for that application you merely need $g$ to be holomorphic in a neighborhood of that circle.
      $endgroup$
      – user98602
      Dec 30 '18 at 22:04














      $begingroup$
      This is a nice answer under the additional assumption that $f$ is entire; however, the OP does not assume that. Your proof uses that assumption twice, once when considering the domain of $g(z)$ via $f(1/z)$ and again when invoking facts about bounded functions.
      $endgroup$
      – Greg Martin
      Dec 31 '18 at 7:33




      $begingroup$
      This is a nice answer under the additional assumption that $f$ is entire; however, the OP does not assume that. Your proof uses that assumption twice, once when considering the domain of $g(z)$ via $f(1/z)$ and again when invoking facts about bounded functions.
      $endgroup$
      – Greg Martin
      Dec 31 '18 at 7:33












      $begingroup$
      @GregMartin It does not use this assumption but I think I didn't explain that clearly enough. One sees that on a neighborhood of $S^1$ contained in the domain of $f$, that $g$ is holomorphic in that neighood and vanishes along the circle; so $f(1/z) = 1/f(z)$ in that nbhd. One may naively extend $f$ to the entire complex plane by the rule $f(1/z) = 1/f(z)$ for $|z| > 1$. Because this formula is true near the circle (where $f$ is holomorphic), and both the given def'ns for $f$ on the int. and ext. of the circle are holomorphic, we see that $f$ has a canonical extension to an entire function.
      $endgroup$
      – user98602
      Dec 31 '18 at 7:38




      $begingroup$
      @GregMartin It does not use this assumption but I think I didn't explain that clearly enough. One sees that on a neighborhood of $S^1$ contained in the domain of $f$, that $g$ is holomorphic in that neighood and vanishes along the circle; so $f(1/z) = 1/f(z)$ in that nbhd. One may naively extend $f$ to the entire complex plane by the rule $f(1/z) = 1/f(z)$ for $|z| > 1$. Because this formula is true near the circle (where $f$ is holomorphic), and both the given def'ns for $f$ on the int. and ext. of the circle are holomorphic, we see that $f$ has a canonical extension to an entire function.
      $endgroup$
      – user98602
      Dec 31 '18 at 7:38












      $begingroup$
      (It was precisely in obtaining the entire extension that we used that $f$ was holomorphic on the entire disc.) As a side comment, I originally wanted to use the Schwarz reflection principle, since $f(S^1) subset S^1$; that also shows that $f$ has a holomorphic extension to all of $Bbb C$. But then I saw there was a more elementary way to see that in this case.
      $endgroup$
      – user98602
      Dec 31 '18 at 7:45






      $begingroup$
      (It was precisely in obtaining the entire extension that we used that $f$ was holomorphic on the entire disc.) As a side comment, I originally wanted to use the Schwarz reflection principle, since $f(S^1) subset S^1$; that also shows that $f$ has a holomorphic extension to all of $Bbb C$. But then I saw there was a more elementary way to see that in this case.
      $endgroup$
      – user98602
      Dec 31 '18 at 7:45













      3












      $begingroup$

      Sequence of hints:




      • If $f$ and $g$ are two such functions, show they must be equal.

      • If $f(z)$ is such a function, show that $g(z) = overline{f(bar z)}$ (its "reflection in the real axis") is another such function.

      • If $f(z)$ is such a function, show that $f(x)$ is real-valued for all $xin(-1,1)$.

      • If $f(z)$ is such a function, show that $h(z) = overline{f(-bar z)}$ (its "reflection in the imaginary axis") is another such function.

      • If $f(z)$ is such a function, show that $f(x)=f(-x)$ for all $xin(-1,1)$.






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        Sequence of hints:




        • If $f$ and $g$ are two such functions, show they must be equal.

        • If $f(z)$ is such a function, show that $g(z) = overline{f(bar z)}$ (its "reflection in the real axis") is another such function.

        • If $f(z)$ is such a function, show that $f(x)$ is real-valued for all $xin(-1,1)$.

        • If $f(z)$ is such a function, show that $h(z) = overline{f(-bar z)}$ (its "reflection in the imaginary axis") is another such function.

        • If $f(z)$ is such a function, show that $f(x)=f(-x)$ for all $xin(-1,1)$.






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          Sequence of hints:




          • If $f$ and $g$ are two such functions, show they must be equal.

          • If $f(z)$ is such a function, show that $g(z) = overline{f(bar z)}$ (its "reflection in the real axis") is another such function.

          • If $f(z)$ is such a function, show that $f(x)$ is real-valued for all $xin(-1,1)$.

          • If $f(z)$ is such a function, show that $h(z) = overline{f(-bar z)}$ (its "reflection in the imaginary axis") is another such function.

          • If $f(z)$ is such a function, show that $f(x)=f(-x)$ for all $xin(-1,1)$.






          share|cite|improve this answer









          $endgroup$



          Sequence of hints:




          • If $f$ and $g$ are two such functions, show they must be equal.

          • If $f(z)$ is such a function, show that $g(z) = overline{f(bar z)}$ (its "reflection in the real axis") is another such function.

          • If $f(z)$ is such a function, show that $f(x)$ is real-valued for all $xin(-1,1)$.

          • If $f(z)$ is such a function, show that $h(z) = overline{f(-bar z)}$ (its "reflection in the imaginary axis") is another such function.

          • If $f(z)$ is such a function, show that $f(x)=f(-x)$ for all $xin(-1,1)$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 30 '18 at 21:31









          Greg MartinGreg Martin

          35.6k23464




          35.6k23464























              3












              $begingroup$

              Nope. $$oint_{|z|=1} e^{iIm z} dz = iint_0^{2pi} e^{i(sintheta + theta)} dtheta = -2pi iJ_1(1) ne 0$$






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                Nope. $$oint_{|z|=1} e^{iIm z} dz = iint_0^{2pi} e^{i(sintheta + theta)} dtheta = -2pi iJ_1(1) ne 0$$






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  Nope. $$oint_{|z|=1} e^{iIm z} dz = iint_0^{2pi} e^{i(sintheta + theta)} dtheta = -2pi iJ_1(1) ne 0$$






                  share|cite|improve this answer









                  $endgroup$



                  Nope. $$oint_{|z|=1} e^{iIm z} dz = iint_0^{2pi} e^{i(sintheta + theta)} dtheta = -2pi iJ_1(1) ne 0$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 30 '18 at 21:34









                  achille huiachille hui

                  96.1k5132260




                  96.1k5132260






























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