Proof that, if $f$ and $|f|$ are analytic, then $f$ is constant












1












$begingroup$


I am trying to show that assuming for an analytic $f$ we have $|f|$ is also analytic. That the original f must be constant.



My original thought was to use the Cauchy Riemann equations to try and show that the partial derivatives are equal to their negatives. I am unsure how to carry this out however, as I am struggling to express the partial derivatives of $|f|$ in terms of the partial derivatives of $f$.



Could someone please let me know if I'm on the right track, or else give me a nudge in the correct direction.



Thanks in advance






complex-analysis analytic-functions







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  • 1




    $begingroup$
    Analytic where? Connectivity has to enter into this somehow.
    $endgroup$
    – zhw.
    Dec 2 '18 at 19:12
















1












$begingroup$


I am trying to show that assuming for an analytic $f$ we have $|f|$ is also analytic. That the original f must be constant.



My original thought was to use the Cauchy Riemann equations to try and show that the partial derivatives are equal to their negatives. I am unsure how to carry this out however, as I am struggling to express the partial derivatives of $|f|$ in terms of the partial derivatives of $f$.



Could someone please let me know if I'm on the right track, or else give me a nudge in the correct direction.



Thanks in advance






complex-analysis analytic-functions







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Analytic where? Connectivity has to enter into this somehow.
    $endgroup$
    – zhw.
    Dec 2 '18 at 19:12














1












1








1





$begingroup$


I am trying to show that assuming for an analytic $f$ we have $|f|$ is also analytic. That the original f must be constant.



My original thought was to use the Cauchy Riemann equations to try and show that the partial derivatives are equal to their negatives. I am unsure how to carry this out however, as I am struggling to express the partial derivatives of $|f|$ in terms of the partial derivatives of $f$.



Could someone please let me know if I'm on the right track, or else give me a nudge in the correct direction.



Thanks in advance






complex-analysis analytic-functions







share|cite|improve this question











$endgroup$




I am trying to show that assuming for an analytic $f$ we have $|f|$ is also analytic. That the original f must be constant.



My original thought was to use the Cauchy Riemann equations to try and show that the partial derivatives are equal to their negatives. I am unsure how to carry this out however, as I am struggling to express the partial derivatives of $|f|$ in terms of the partial derivatives of $f$.



Could someone please let me know if I'm on the right track, or else give me a nudge in the correct direction.



Thanks in advance






complex-analysis analytic-functions




complex-analysis analytic-functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 2 '18 at 19:36









Did

247k23222457




247k23222457










asked Dec 2 '18 at 16:14









user601175user601175

114




114








  • 1




    $begingroup$
    Analytic where? Connectivity has to enter into this somehow.
    $endgroup$
    – zhw.
    Dec 2 '18 at 19:12














  • 1




    $begingroup$
    Analytic where? Connectivity has to enter into this somehow.
    $endgroup$
    – zhw.
    Dec 2 '18 at 19:12








1




1




$begingroup$
Analytic where? Connectivity has to enter into this somehow.
$endgroup$
– zhw.
Dec 2 '18 at 19:12




$begingroup$
Analytic where? Connectivity has to enter into this somehow.
$endgroup$
– zhw.
Dec 2 '18 at 19:12










3 Answers
3






active

oldest

votes


















0












$begingroup$

$|f|=|u+iv|=sqrt{u^2+v^2}$ is analytic, $u=u(x,y), v=v(x,y)$.



$implies |f|=U+iV; U=sqrt{u^2+v^2}, V=0$



$U_x=V_yimplies uu_x+vv_x=0$



$U_y=-V_ximplies uu_y+vv_y=0$



From the analyticity of $f$, we have $u_x=v_y, u_y=-v_x;$



$implies uu_x-vu_y=0$



$implies uu_y+vu_x=0$



$implies (u^2+v^2)u_y=(u^2+v^2)u_x=0$



For non-trivial solution, $u^2+v^2neq0$



$implies u_x=u_y=v_x=v_y=0$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I used dyU=-dxV=0 to get udyu+vdxu=0 how do you then get to dyu=dxu=0
    $endgroup$
    – user601175
    Dec 2 '18 at 17:19










  • $begingroup$
    I edited the answer for the complete solution.
    $endgroup$
    – Shubham Johri
    Dec 2 '18 at 17:59



















3












$begingroup$

If $f$ and $|f|$ are both analytic (in the whole complex plane $mathbb C$), $g:=frac{f}{1+|f|}$ is analytic in the whole complex plane and $|g|< 1$. So $g$ is constant $f=c(1+|f|)$ and, in particular, $|f|=|c|(1+|f|)=|c|+|c||f|$ and $|f|=frac{|c|}{1-|c|}$ and you have that $|f|$ is constant. Then, using that $f=c(1+|f|)$, you'll get that $f=cleft(1+frac{|c|}{1-|c|}right)$ which is a constant.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    How do you know that g is constant. Are all analytic functions with modulus <1 constant?
    $endgroup$
    – user601175
    Dec 2 '18 at 17:08






  • 1




    $begingroup$
    Any bounded entire function is constant (Liouville's Theorem)
    $endgroup$
    – Tito Eliatron
    Dec 2 '18 at 17:10



















0












$begingroup$

Assume $f,|f|$ are analytic in the connected open set $U.$ If $fequiv 0,$ we're done. Otherwise, using continuity, there is a disc $D(a,r)subset U$ where $f$ is nonzero. Now $|f|$ analytic implies $|f|^2= fbar f$ is analytic. Thus in $D(a,r),$ $(fbar f)/f = bar f$ is analytic. Now use the CR equations (this is easy) to see $f$ must be constant in $D(a,r),$ hence constant in $U$ by the identity principle.






share|cite|improve this answer









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    3 Answers
    3






    active

    oldest

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    3 Answers
    3






    active

    oldest

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    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    $|f|=|u+iv|=sqrt{u^2+v^2}$ is analytic, $u=u(x,y), v=v(x,y)$.



    $implies |f|=U+iV; U=sqrt{u^2+v^2}, V=0$



    $U_x=V_yimplies uu_x+vv_x=0$



    $U_y=-V_ximplies uu_y+vv_y=0$



    From the analyticity of $f$, we have $u_x=v_y, u_y=-v_x;$



    $implies uu_x-vu_y=0$



    $implies uu_y+vu_x=0$



    $implies (u^2+v^2)u_y=(u^2+v^2)u_x=0$



    For non-trivial solution, $u^2+v^2neq0$



    $implies u_x=u_y=v_x=v_y=0$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I used dyU=-dxV=0 to get udyu+vdxu=0 how do you then get to dyu=dxu=0
      $endgroup$
      – user601175
      Dec 2 '18 at 17:19










    • $begingroup$
      I edited the answer for the complete solution.
      $endgroup$
      – Shubham Johri
      Dec 2 '18 at 17:59
















    0












    $begingroup$

    $|f|=|u+iv|=sqrt{u^2+v^2}$ is analytic, $u=u(x,y), v=v(x,y)$.



    $implies |f|=U+iV; U=sqrt{u^2+v^2}, V=0$



    $U_x=V_yimplies uu_x+vv_x=0$



    $U_y=-V_ximplies uu_y+vv_y=0$



    From the analyticity of $f$, we have $u_x=v_y, u_y=-v_x;$



    $implies uu_x-vu_y=0$



    $implies uu_y+vu_x=0$



    $implies (u^2+v^2)u_y=(u^2+v^2)u_x=0$



    For non-trivial solution, $u^2+v^2neq0$



    $implies u_x=u_y=v_x=v_y=0$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I used dyU=-dxV=0 to get udyu+vdxu=0 how do you then get to dyu=dxu=0
      $endgroup$
      – user601175
      Dec 2 '18 at 17:19










    • $begingroup$
      I edited the answer for the complete solution.
      $endgroup$
      – Shubham Johri
      Dec 2 '18 at 17:59














    0












    0








    0





    $begingroup$

    $|f|=|u+iv|=sqrt{u^2+v^2}$ is analytic, $u=u(x,y), v=v(x,y)$.



    $implies |f|=U+iV; U=sqrt{u^2+v^2}, V=0$



    $U_x=V_yimplies uu_x+vv_x=0$



    $U_y=-V_ximplies uu_y+vv_y=0$



    From the analyticity of $f$, we have $u_x=v_y, u_y=-v_x;$



    $implies uu_x-vu_y=0$



    $implies uu_y+vu_x=0$



    $implies (u^2+v^2)u_y=(u^2+v^2)u_x=0$



    For non-trivial solution, $u^2+v^2neq0$



    $implies u_x=u_y=v_x=v_y=0$






    share|cite|improve this answer











    $endgroup$



    $|f|=|u+iv|=sqrt{u^2+v^2}$ is analytic, $u=u(x,y), v=v(x,y)$.



    $implies |f|=U+iV; U=sqrt{u^2+v^2}, V=0$



    $U_x=V_yimplies uu_x+vv_x=0$



    $U_y=-V_ximplies uu_y+vv_y=0$



    From the analyticity of $f$, we have $u_x=v_y, u_y=-v_x;$



    $implies uu_x-vu_y=0$



    $implies uu_y+vu_x=0$



    $implies (u^2+v^2)u_y=(u^2+v^2)u_x=0$



    For non-trivial solution, $u^2+v^2neq0$



    $implies u_x=u_y=v_x=v_y=0$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 2 '18 at 18:06

























    answered Dec 2 '18 at 16:35









    Shubham JohriShubham Johri

    4,759717




    4,759717












    • $begingroup$
      I used dyU=-dxV=0 to get udyu+vdxu=0 how do you then get to dyu=dxu=0
      $endgroup$
      – user601175
      Dec 2 '18 at 17:19










    • $begingroup$
      I edited the answer for the complete solution.
      $endgroup$
      – Shubham Johri
      Dec 2 '18 at 17:59


















    • $begingroup$
      I used dyU=-dxV=0 to get udyu+vdxu=0 how do you then get to dyu=dxu=0
      $endgroup$
      – user601175
      Dec 2 '18 at 17:19










    • $begingroup$
      I edited the answer for the complete solution.
      $endgroup$
      – Shubham Johri
      Dec 2 '18 at 17:59
















    $begingroup$
    I used dyU=-dxV=0 to get udyu+vdxu=0 how do you then get to dyu=dxu=0
    $endgroup$
    – user601175
    Dec 2 '18 at 17:19




    $begingroup$
    I used dyU=-dxV=0 to get udyu+vdxu=0 how do you then get to dyu=dxu=0
    $endgroup$
    – user601175
    Dec 2 '18 at 17:19












    $begingroup$
    I edited the answer for the complete solution.
    $endgroup$
    – Shubham Johri
    Dec 2 '18 at 17:59




    $begingroup$
    I edited the answer for the complete solution.
    $endgroup$
    – Shubham Johri
    Dec 2 '18 at 17:59











    3












    $begingroup$

    If $f$ and $|f|$ are both analytic (in the whole complex plane $mathbb C$), $g:=frac{f}{1+|f|}$ is analytic in the whole complex plane and $|g|< 1$. So $g$ is constant $f=c(1+|f|)$ and, in particular, $|f|=|c|(1+|f|)=|c|+|c||f|$ and $|f|=frac{|c|}{1-|c|}$ and you have that $|f|$ is constant. Then, using that $f=c(1+|f|)$, you'll get that $f=cleft(1+frac{|c|}{1-|c|}right)$ which is a constant.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      How do you know that g is constant. Are all analytic functions with modulus <1 constant?
      $endgroup$
      – user601175
      Dec 2 '18 at 17:08






    • 1




      $begingroup$
      Any bounded entire function is constant (Liouville's Theorem)
      $endgroup$
      – Tito Eliatron
      Dec 2 '18 at 17:10
















    3












    $begingroup$

    If $f$ and $|f|$ are both analytic (in the whole complex plane $mathbb C$), $g:=frac{f}{1+|f|}$ is analytic in the whole complex plane and $|g|< 1$. So $g$ is constant $f=c(1+|f|)$ and, in particular, $|f|=|c|(1+|f|)=|c|+|c||f|$ and $|f|=frac{|c|}{1-|c|}$ and you have that $|f|$ is constant. Then, using that $f=c(1+|f|)$, you'll get that $f=cleft(1+frac{|c|}{1-|c|}right)$ which is a constant.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      How do you know that g is constant. Are all analytic functions with modulus <1 constant?
      $endgroup$
      – user601175
      Dec 2 '18 at 17:08






    • 1




      $begingroup$
      Any bounded entire function is constant (Liouville's Theorem)
      $endgroup$
      – Tito Eliatron
      Dec 2 '18 at 17:10














    3












    3








    3





    $begingroup$

    If $f$ and $|f|$ are both analytic (in the whole complex plane $mathbb C$), $g:=frac{f}{1+|f|}$ is analytic in the whole complex plane and $|g|< 1$. So $g$ is constant $f=c(1+|f|)$ and, in particular, $|f|=|c|(1+|f|)=|c|+|c||f|$ and $|f|=frac{|c|}{1-|c|}$ and you have that $|f|$ is constant. Then, using that $f=c(1+|f|)$, you'll get that $f=cleft(1+frac{|c|}{1-|c|}right)$ which is a constant.






    share|cite|improve this answer









    $endgroup$



    If $f$ and $|f|$ are both analytic (in the whole complex plane $mathbb C$), $g:=frac{f}{1+|f|}$ is analytic in the whole complex plane and $|g|< 1$. So $g$ is constant $f=c(1+|f|)$ and, in particular, $|f|=|c|(1+|f|)=|c|+|c||f|$ and $|f|=frac{|c|}{1-|c|}$ and you have that $|f|$ is constant. Then, using that $f=c(1+|f|)$, you'll get that $f=cleft(1+frac{|c|}{1-|c|}right)$ which is a constant.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 2 '18 at 16:35









    Tito EliatronTito Eliatron

    1,448622




    1,448622












    • $begingroup$
      How do you know that g is constant. Are all analytic functions with modulus <1 constant?
      $endgroup$
      – user601175
      Dec 2 '18 at 17:08






    • 1




      $begingroup$
      Any bounded entire function is constant (Liouville's Theorem)
      $endgroup$
      – Tito Eliatron
      Dec 2 '18 at 17:10


















    • $begingroup$
      How do you know that g is constant. Are all analytic functions with modulus <1 constant?
      $endgroup$
      – user601175
      Dec 2 '18 at 17:08






    • 1




      $begingroup$
      Any bounded entire function is constant (Liouville's Theorem)
      $endgroup$
      – Tito Eliatron
      Dec 2 '18 at 17:10
















    $begingroup$
    How do you know that g is constant. Are all analytic functions with modulus <1 constant?
    $endgroup$
    – user601175
    Dec 2 '18 at 17:08




    $begingroup$
    How do you know that g is constant. Are all analytic functions with modulus <1 constant?
    $endgroup$
    – user601175
    Dec 2 '18 at 17:08




    1




    1




    $begingroup$
    Any bounded entire function is constant (Liouville's Theorem)
    $endgroup$
    – Tito Eliatron
    Dec 2 '18 at 17:10




    $begingroup$
    Any bounded entire function is constant (Liouville's Theorem)
    $endgroup$
    – Tito Eliatron
    Dec 2 '18 at 17:10











    0












    $begingroup$

    Assume $f,|f|$ are analytic in the connected open set $U.$ If $fequiv 0,$ we're done. Otherwise, using continuity, there is a disc $D(a,r)subset U$ where $f$ is nonzero. Now $|f|$ analytic implies $|f|^2= fbar f$ is analytic. Thus in $D(a,r),$ $(fbar f)/f = bar f$ is analytic. Now use the CR equations (this is easy) to see $f$ must be constant in $D(a,r),$ hence constant in $U$ by the identity principle.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Assume $f,|f|$ are analytic in the connected open set $U.$ If $fequiv 0,$ we're done. Otherwise, using continuity, there is a disc $D(a,r)subset U$ where $f$ is nonzero. Now $|f|$ analytic implies $|f|^2= fbar f$ is analytic. Thus in $D(a,r),$ $(fbar f)/f = bar f$ is analytic. Now use the CR equations (this is easy) to see $f$ must be constant in $D(a,r),$ hence constant in $U$ by the identity principle.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Assume $f,|f|$ are analytic in the connected open set $U.$ If $fequiv 0,$ we're done. Otherwise, using continuity, there is a disc $D(a,r)subset U$ where $f$ is nonzero. Now $|f|$ analytic implies $|f|^2= fbar f$ is analytic. Thus in $D(a,r),$ $(fbar f)/f = bar f$ is analytic. Now use the CR equations (this is easy) to see $f$ must be constant in $D(a,r),$ hence constant in $U$ by the identity principle.






        share|cite|improve this answer









        $endgroup$



        Assume $f,|f|$ are analytic in the connected open set $U.$ If $fequiv 0,$ we're done. Otherwise, using continuity, there is a disc $D(a,r)subset U$ where $f$ is nonzero. Now $|f|$ analytic implies $|f|^2= fbar f$ is analytic. Thus in $D(a,r),$ $(fbar f)/f = bar f$ is analytic. Now use the CR equations (this is easy) to see $f$ must be constant in $D(a,r),$ hence constant in $U$ by the identity principle.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 2 '18 at 19:33









        zhw.zhw.

        71.9k43075




        71.9k43075






























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