Existence of analytic function having different values on $|z|=1$ and $|z|=2$












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$begingroup$


Does there exist an entire function satisying $f(z)=z$ for $|z|=1$ and $f(z)=z^2$ for $|z|=2$?



I think no, but what argument would do here? Would it be maximum modulus or Liouville theorem? A counterexample should involve power series, I think? Any hints? Thanks beforehand.






complex-analysis entire-functions







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$endgroup$

















    1












    $begingroup$


    Does there exist an entire function satisying $f(z)=z$ for $|z|=1$ and $f(z)=z^2$ for $|z|=2$?



    I think no, but what argument would do here? Would it be maximum modulus or Liouville theorem? A counterexample should involve power series, I think? Any hints? Thanks beforehand.






    complex-analysis entire-functions







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Does there exist an entire function satisying $f(z)=z$ for $|z|=1$ and $f(z)=z^2$ for $|z|=2$?



      I think no, but what argument would do here? Would it be maximum modulus or Liouville theorem? A counterexample should involve power series, I think? Any hints? Thanks beforehand.






      complex-analysis entire-functions







      share|cite|improve this question









      $endgroup$




      Does there exist an entire function satisying $f(z)=z$ for $|z|=1$ and $f(z)=z^2$ for $|z|=2$?



      I think no, but what argument would do here? Would it be maximum modulus or Liouville theorem? A counterexample should involve power series, I think? Any hints? Thanks beforehand.






      complex-analysis entire-functions




      complex-analysis entire-functions






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 25 at 9:02









      vidyarthividyarthi

      3,0361833




      3,0361833






















          2 Answers
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          5












          $begingroup$

          If $f(z)=z$ for $|z|=1$ then $f(z)-z$ is an entire function whose zeros have a limit point. This implies $f(z)=z$ for all $z$. So no such function can exist.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            oh! what an easy argument. Just missed it though
            $endgroup$
            – vidyarthi
            Jan 25 at 9:05



















          1












          $begingroup$

          We have



          $$f'(0)= frac{1}{2 pi i}int_{|z|=1} frac{f(w)}{w^2} dw =frac{1}{2 pi i}int_{|z|=1} frac{w}{w^2} dw =frac{1}{2 pi i}int_{|z|=1} frac{1}{w} dw=1$$



          and



          $$f'(0)= frac{1}{2 pi i}int_{|z|=2} frac{f(w)}{w^2} dw =frac{1}{2 pi i}int_{|z|=2} frac{w^2}{w^2} dw =frac{1}{2 pi i}int_{|z|=2} 1 dw=0.$$



          This contradiction shows that no such function exists.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            that's the power of complex integration
            $endgroup$
            – vidyarthi
            Jan 25 at 11:10











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          2 Answers
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          active

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          2 Answers
          2






          active

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          active

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          active

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          5












          $begingroup$

          If $f(z)=z$ for $|z|=1$ then $f(z)-z$ is an entire function whose zeros have a limit point. This implies $f(z)=z$ for all $z$. So no such function can exist.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            oh! what an easy argument. Just missed it though
            $endgroup$
            – vidyarthi
            Jan 25 at 9:05
















          5












          $begingroup$

          If $f(z)=z$ for $|z|=1$ then $f(z)-z$ is an entire function whose zeros have a limit point. This implies $f(z)=z$ for all $z$. So no such function can exist.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            oh! what an easy argument. Just missed it though
            $endgroup$
            – vidyarthi
            Jan 25 at 9:05














          5












          5








          5





          $begingroup$

          If $f(z)=z$ for $|z|=1$ then $f(z)-z$ is an entire function whose zeros have a limit point. This implies $f(z)=z$ for all $z$. So no such function can exist.






          share|cite|improve this answer









          $endgroup$



          If $f(z)=z$ for $|z|=1$ then $f(z)-z$ is an entire function whose zeros have a limit point. This implies $f(z)=z$ for all $z$. So no such function can exist.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 25 at 9:04









          Kavi Rama MurthyKavi Rama Murthy

          68.3k53069




          68.3k53069












          • $begingroup$
            oh! what an easy argument. Just missed it though
            $endgroup$
            – vidyarthi
            Jan 25 at 9:05


















          • $begingroup$
            oh! what an easy argument. Just missed it though
            $endgroup$
            – vidyarthi
            Jan 25 at 9:05
















          $begingroup$
          oh! what an easy argument. Just missed it though
          $endgroup$
          – vidyarthi
          Jan 25 at 9:05




          $begingroup$
          oh! what an easy argument. Just missed it though
          $endgroup$
          – vidyarthi
          Jan 25 at 9:05











          1












          $begingroup$

          We have



          $$f'(0)= frac{1}{2 pi i}int_{|z|=1} frac{f(w)}{w^2} dw =frac{1}{2 pi i}int_{|z|=1} frac{w}{w^2} dw =frac{1}{2 pi i}int_{|z|=1} frac{1}{w} dw=1$$



          and



          $$f'(0)= frac{1}{2 pi i}int_{|z|=2} frac{f(w)}{w^2} dw =frac{1}{2 pi i}int_{|z|=2} frac{w^2}{w^2} dw =frac{1}{2 pi i}int_{|z|=2} 1 dw=0.$$



          This contradiction shows that no such function exists.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            that's the power of complex integration
            $endgroup$
            – vidyarthi
            Jan 25 at 11:10
















          1












          $begingroup$

          We have



          $$f'(0)= frac{1}{2 pi i}int_{|z|=1} frac{f(w)}{w^2} dw =frac{1}{2 pi i}int_{|z|=1} frac{w}{w^2} dw =frac{1}{2 pi i}int_{|z|=1} frac{1}{w} dw=1$$



          and



          $$f'(0)= frac{1}{2 pi i}int_{|z|=2} frac{f(w)}{w^2} dw =frac{1}{2 pi i}int_{|z|=2} frac{w^2}{w^2} dw =frac{1}{2 pi i}int_{|z|=2} 1 dw=0.$$



          This contradiction shows that no such function exists.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            that's the power of complex integration
            $endgroup$
            – vidyarthi
            Jan 25 at 11:10














          1












          1








          1





          $begingroup$

          We have



          $$f'(0)= frac{1}{2 pi i}int_{|z|=1} frac{f(w)}{w^2} dw =frac{1}{2 pi i}int_{|z|=1} frac{w}{w^2} dw =frac{1}{2 pi i}int_{|z|=1} frac{1}{w} dw=1$$



          and



          $$f'(0)= frac{1}{2 pi i}int_{|z|=2} frac{f(w)}{w^2} dw =frac{1}{2 pi i}int_{|z|=2} frac{w^2}{w^2} dw =frac{1}{2 pi i}int_{|z|=2} 1 dw=0.$$



          This contradiction shows that no such function exists.






          share|cite|improve this answer









          $endgroup$



          We have



          $$f'(0)= frac{1}{2 pi i}int_{|z|=1} frac{f(w)}{w^2} dw =frac{1}{2 pi i}int_{|z|=1} frac{w}{w^2} dw =frac{1}{2 pi i}int_{|z|=1} frac{1}{w} dw=1$$



          and



          $$f'(0)= frac{1}{2 pi i}int_{|z|=2} frac{f(w)}{w^2} dw =frac{1}{2 pi i}int_{|z|=2} frac{w^2}{w^2} dw =frac{1}{2 pi i}int_{|z|=2} 1 dw=0.$$



          This contradiction shows that no such function exists.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 25 at 11:01









          FredFred

          48.5k11849




          48.5k11849












          • $begingroup$
            that's the power of complex integration
            $endgroup$
            – vidyarthi
            Jan 25 at 11:10


















          • $begingroup$
            that's the power of complex integration
            $endgroup$
            – vidyarthi
            Jan 25 at 11:10
















          $begingroup$
          that's the power of complex integration
          $endgroup$
          – vidyarthi
          Jan 25 at 11:10




          $begingroup$
          that's the power of complex integration
          $endgroup$
          – vidyarthi
          Jan 25 at 11:10


















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