Proving a function is constant, if a sum of the real and imaginary parts is bounded











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Let $g$ be an entire function such that there exist three
real numbers $a$, $b$, $c$ such that $a$ and $b$ are not both $0$ and such that
$$a operatorname{Re}(g(z)) + b operatorname{Im}(g(z)) ≤ c$$
for all $ z ∈ mathbb{C}$.



Show that $g$ is constant.




Solving simpler versions of this problem involves defining a new function, say $h(z)$, that is some (often exponential) function we can find to be bounded. What do we do in our case?










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    Let $g$ be an entire function such that there exist three
    real numbers $a$, $b$, $c$ such that $a$ and $b$ are not both $0$ and such that
    $$a operatorname{Re}(g(z)) + b operatorname{Im}(g(z)) ≤ c$$
    for all $ z ∈ mathbb{C}$.



    Show that $g$ is constant.




    Solving simpler versions of this problem involves defining a new function, say $h(z)$, that is some (often exponential) function we can find to be bounded. What do we do in our case?










    share|cite|improve this question


























      up vote
      0
      down vote

      favorite
      1









      up vote
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      down vote

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      1






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      Let $g$ be an entire function such that there exist three
      real numbers $a$, $b$, $c$ such that $a$ and $b$ are not both $0$ and such that
      $$a operatorname{Re}(g(z)) + b operatorname{Im}(g(z)) ≤ c$$
      for all $ z ∈ mathbb{C}$.



      Show that $g$ is constant.




      Solving simpler versions of this problem involves defining a new function, say $h(z)$, that is some (often exponential) function we can find to be bounded. What do we do in our case?










      share|cite|improve this question
















      Let $g$ be an entire function such that there exist three
      real numbers $a$, $b$, $c$ such that $a$ and $b$ are not both $0$ and such that
      $$a operatorname{Re}(g(z)) + b operatorname{Im}(g(z)) ≤ c$$
      for all $ z ∈ mathbb{C}$.



      Show that $g$ is constant.




      Solving simpler versions of this problem involves defining a new function, say $h(z)$, that is some (often exponential) function we can find to be bounded. What do we do in our case?







      complex-analysis






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      share|cite|improve this question













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      edited Nov 22 at 18:42









      egreg

      176k1484198




      176k1484198










      asked Nov 22 at 18:31









      Dino

      836




      836






















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          Let $h(z)=(a-bi)g(z)$. Then$$(forall zinmathbb{C}):operatorname{Re}h(z)leqslant c.$$Now, define $f(z)=e^{h(z)}$. Then$$(forall zinmathbb{C}):bigllvert f(z)bigrrvert=e^{operatorname{Re}h(z)}leqslant e^c.$$Can you take it from here?






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            up vote
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            down vote



            accepted










            Let $h(z)=(a-bi)g(z)$. Then$$(forall zinmathbb{C}):operatorname{Re}h(z)leqslant c.$$Now, define $f(z)=e^{h(z)}$. Then$$(forall zinmathbb{C}):bigllvert f(z)bigrrvert=e^{operatorname{Re}h(z)}leqslant e^c.$$Can you take it from here?






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              up vote
              2
              down vote



              accepted










              Let $h(z)=(a-bi)g(z)$. Then$$(forall zinmathbb{C}):operatorname{Re}h(z)leqslant c.$$Now, define $f(z)=e^{h(z)}$. Then$$(forall zinmathbb{C}):bigllvert f(z)bigrrvert=e^{operatorname{Re}h(z)}leqslant e^c.$$Can you take it from here?






              share|cite|improve this answer























                up vote
                2
                down vote



                accepted







                up vote
                2
                down vote



                accepted






                Let $h(z)=(a-bi)g(z)$. Then$$(forall zinmathbb{C}):operatorname{Re}h(z)leqslant c.$$Now, define $f(z)=e^{h(z)}$. Then$$(forall zinmathbb{C}):bigllvert f(z)bigrrvert=e^{operatorname{Re}h(z)}leqslant e^c.$$Can you take it from here?






                share|cite|improve this answer












                Let $h(z)=(a-bi)g(z)$. Then$$(forall zinmathbb{C}):operatorname{Re}h(z)leqslant c.$$Now, define $f(z)=e^{h(z)}$. Then$$(forall zinmathbb{C}):bigllvert f(z)bigrrvert=e^{operatorname{Re}h(z)}leqslant e^c.$$Can you take it from here?







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 22 at 18:39









                José Carlos Santos

                147k22117217




                147k22117217






























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