All Entire Functions $f(f(z))=f(z)$

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I have read the thread but it was hard for me to understand, I have tried to to bound the function so it will be constant and to think about function other the the identity the fulfil the condition, can you please explain me how to approach this?




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    I have read the thread but it was hard for me to understand, I have tried to to bound the function so it will be constant and to think about function other the the identity the fulfil the condition, can you please explain me how to approach this?




    complex-analysis




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      I have read the thread but it was hard for me to understand, I have tried to to bound the function so it will be constant and to think about function other the the identity the fulfil the condition, can you please explain me how to approach this?




      complex-analysis




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      I have read the thread but it was hard for me to understand, I have tried to to bound the function so it will be constant and to think about function other the the identity the fulfil the condition, can you please explain me how to approach this?





      complex-analysis





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      asked Jul 16 at 19:16









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          If $f$ is a nonconstant entire function, then the image of $f$ is open, so it contains a bounded infinite subset along with the limit points. Note that $f$ is the identity map on its image, so the Identity Theorem implies that $f$ is the identity function. If $f$ is constant, then there is nothing to prove.






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            If $f$ is a nonconstant entire function, then the image of $f$ is open, so it contains a bounded infinite subset along with the limit points. Note that $f$ is the identity map on its image, so the Identity Theorem implies that $f$ is the identity function. If $f$ is constant, then there is nothing to prove.






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              If $f$ is a nonconstant entire function, then the image of $f$ is open, so it contains a bounded infinite subset along with the limit points. Note that $f$ is the identity map on its image, so the Identity Theorem implies that $f$ is the identity function. If $f$ is constant, then there is nothing to prove.






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                up vote
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                If $f$ is a nonconstant entire function, then the image of $f$ is open, so it contains a bounded infinite subset along with the limit points. Note that $f$ is the identity map on its image, so the Identity Theorem implies that $f$ is the identity function. If $f$ is constant, then there is nothing to prove.






                share|cite|improve this answer













                If $f$ is a nonconstant entire function, then the image of $f$ is open, so it contains a bounded infinite subset along with the limit points. Note that $f$ is the identity map on its image, so the Identity Theorem implies that $f$ is the identity function. If $f$ is constant, then there is nothing to prove.







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                answered Jul 16 at 19:23









                Batominovski

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