Is there is an entire function $f$ such that $f(1)=pi$ and $f'(z)=|z|f(z)$ for all $zinmathbb C$?












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Problem: Is there is an entire function $f$ such that $f(1)=pi$ and $f'(z)=|z|f(z)$ for all $zinmathbb C$?




Question: Is the following correct?



On the unit circle line $partial K_1(0)$ the above equation is equivalent to $f'(z)=f(z)$. According to Picard-Lindelof the IVP with respect to $f(1)=pi$ is uniquely solved by $f(z)=frac{pi}{e}e^z$. Since $partial K_1(0)$ has an accumulation point the identity theorem can be applied and $f(z)=frac{pi}{e}e^z$ on the entire $mathbb C$-plane. This however contradicts $f'(z)=|z|f(z)$ on each point outside of the unit circle line.






complex-analysis proof-verification entire-functions







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$endgroup$

















    0












    $begingroup$



    Problem: Is there is an entire function $f$ such that $f(1)=pi$ and $f'(z)=|z|f(z)$ for all $zinmathbb C$?




    Question: Is the following correct?



    On the unit circle line $partial K_1(0)$ the above equation is equivalent to $f'(z)=f(z)$. According to Picard-Lindelof the IVP with respect to $f(1)=pi$ is uniquely solved by $f(z)=frac{pi}{e}e^z$. Since $partial K_1(0)$ has an accumulation point the identity theorem can be applied and $f(z)=frac{pi}{e}e^z$ on the entire $mathbb C$-plane. This however contradicts $f'(z)=|z|f(z)$ on each point outside of the unit circle line.






    complex-analysis proof-verification entire-functions







    share|cite|improve this question











    $endgroup$















      0












      0








      0


      1



      $begingroup$



      Problem: Is there is an entire function $f$ such that $f(1)=pi$ and $f'(z)=|z|f(z)$ for all $zinmathbb C$?




      Question: Is the following correct?



      On the unit circle line $partial K_1(0)$ the above equation is equivalent to $f'(z)=f(z)$. According to Picard-Lindelof the IVP with respect to $f(1)=pi$ is uniquely solved by $f(z)=frac{pi}{e}e^z$. Since $partial K_1(0)$ has an accumulation point the identity theorem can be applied and $f(z)=frac{pi}{e}e^z$ on the entire $mathbb C$-plane. This however contradicts $f'(z)=|z|f(z)$ on each point outside of the unit circle line.






      complex-analysis proof-verification entire-functions







      share|cite|improve this question











      $endgroup$





      Problem: Is there is an entire function $f$ such that $f(1)=pi$ and $f'(z)=|z|f(z)$ for all $zinmathbb C$?




      Question: Is the following correct?



      On the unit circle line $partial K_1(0)$ the above equation is equivalent to $f'(z)=f(z)$. According to Picard-Lindelof the IVP with respect to $f(1)=pi$ is uniquely solved by $f(z)=frac{pi}{e}e^z$. Since $partial K_1(0)$ has an accumulation point the identity theorem can be applied and $f(z)=frac{pi}{e}e^z$ on the entire $mathbb C$-plane. This however contradicts $f'(z)=|z|f(z)$ on each point outside of the unit circle line.






      complex-analysis proof-verification entire-functions




      complex-analysis proof-verification entire-functions






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 5 at 11:43









      José Carlos Santos

      165k22132235




      165k22132235










      asked Jan 5 at 11:04









      RedLanternRedLantern

      415




      415






















          1 Answer
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          active

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          3












          $begingroup$

          It's much easier to do it as follows: the set $Z$ of zeros of $f$ is a closed and discrete subset of $mathbb C$. So, on $mathbb{C}setminus Z$, we would have $lvert zrvert=frac{f'(z)}{f(z)}$. But $zmapstolvert zrvert$ is nowhere differentiable.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            So... is my argumentation correct or not?
            $endgroup$
            – RedLantern
            Jan 5 at 11:45






          • 1




            $begingroup$
            It looks fine to me.
            $endgroup$
            – José Carlos Santos
            Jan 5 at 11:49










          • $begingroup$
            Okay, thanks a lot!
            $endgroup$
            – RedLantern
            Jan 5 at 12:01











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          1 Answer
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          active

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          1 Answer
          1






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

          votes









          3












          $begingroup$

          It's much easier to do it as follows: the set $Z$ of zeros of $f$ is a closed and discrete subset of $mathbb C$. So, on $mathbb{C}setminus Z$, we would have $lvert zrvert=frac{f'(z)}{f(z)}$. But $zmapstolvert zrvert$ is nowhere differentiable.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            So... is my argumentation correct or not?
            $endgroup$
            – RedLantern
            Jan 5 at 11:45






          • 1




            $begingroup$
            It looks fine to me.
            $endgroup$
            – José Carlos Santos
            Jan 5 at 11:49










          • $begingroup$
            Okay, thanks a lot!
            $endgroup$
            – RedLantern
            Jan 5 at 12:01
















          3












          $begingroup$

          It's much easier to do it as follows: the set $Z$ of zeros of $f$ is a closed and discrete subset of $mathbb C$. So, on $mathbb{C}setminus Z$, we would have $lvert zrvert=frac{f'(z)}{f(z)}$. But $zmapstolvert zrvert$ is nowhere differentiable.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            So... is my argumentation correct or not?
            $endgroup$
            – RedLantern
            Jan 5 at 11:45






          • 1




            $begingroup$
            It looks fine to me.
            $endgroup$
            – José Carlos Santos
            Jan 5 at 11:49










          • $begingroup$
            Okay, thanks a lot!
            $endgroup$
            – RedLantern
            Jan 5 at 12:01














          3












          3








          3





          $begingroup$

          It's much easier to do it as follows: the set $Z$ of zeros of $f$ is a closed and discrete subset of $mathbb C$. So, on $mathbb{C}setminus Z$, we would have $lvert zrvert=frac{f'(z)}{f(z)}$. But $zmapstolvert zrvert$ is nowhere differentiable.






          share|cite|improve this answer











          $endgroup$



          It's much easier to do it as follows: the set $Z$ of zeros of $f$ is a closed and discrete subset of $mathbb C$. So, on $mathbb{C}setminus Z$, we would have $lvert zrvert=frac{f'(z)}{f(z)}$. But $zmapstolvert zrvert$ is nowhere differentiable.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 5 at 11:47

























          answered Jan 5 at 11:11









          José Carlos SantosJosé Carlos Santos

          165k22132235




          165k22132235








          • 1




            $begingroup$
            So... is my argumentation correct or not?
            $endgroup$
            – RedLantern
            Jan 5 at 11:45






          • 1




            $begingroup$
            It looks fine to me.
            $endgroup$
            – José Carlos Santos
            Jan 5 at 11:49










          • $begingroup$
            Okay, thanks a lot!
            $endgroup$
            – RedLantern
            Jan 5 at 12:01














          • 1




            $begingroup$
            So... is my argumentation correct or not?
            $endgroup$
            – RedLantern
            Jan 5 at 11:45






          • 1




            $begingroup$
            It looks fine to me.
            $endgroup$
            – José Carlos Santos
            Jan 5 at 11:49










          • $begingroup$
            Okay, thanks a lot!
            $endgroup$
            – RedLantern
            Jan 5 at 12:01








          1




          1




          $begingroup$
          So... is my argumentation correct or not?
          $endgroup$
          – RedLantern
          Jan 5 at 11:45




          $begingroup$
          So... is my argumentation correct or not?
          $endgroup$
          – RedLantern
          Jan 5 at 11:45




          1




          1




          $begingroup$
          It looks fine to me.
          $endgroup$
          – José Carlos Santos
          Jan 5 at 11:49




          $begingroup$
          It looks fine to me.
          $endgroup$
          – José Carlos Santos
          Jan 5 at 11:49












          $begingroup$
          Okay, thanks a lot!
          $endgroup$
          – RedLantern
          Jan 5 at 12:01




          $begingroup$
          Okay, thanks a lot!
          $endgroup$
          – RedLantern
          Jan 5 at 12:01


















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