Let $f(x)$ and $g(x)$ are two complex polynomials such that $f^{-1}(c_{i})=g^{-1}(c_{i})$












1












$begingroup$


Let $f(x)$ and $g(x)$ are two complex polynomials such that $f^{-1}(c_{i})=g^{-1}(c_{i})$ for two distinct complex numbers $c_{i}$, $i=1,2$.
Then can we say $f=g$?



Here nothing is given about the multiplicity $f^{-1}(c_{i})$. So any hint please..



Thank you.






complex-analysis polynomials







share|cite|improve this question











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    1












    $begingroup$


    Let $f(x)$ and $g(x)$ are two complex polynomials such that $f^{-1}(c_{i})=g^{-1}(c_{i})$ for two distinct complex numbers $c_{i}$, $i=1,2$.
    Then can we say $f=g$?



    Here nothing is given about the multiplicity $f^{-1}(c_{i})$. So any hint please..



    Thank you.






    complex-analysis polynomials







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Let $f(x)$ and $g(x)$ are two complex polynomials such that $f^{-1}(c_{i})=g^{-1}(c_{i})$ for two distinct complex numbers $c_{i}$, $i=1,2$.
      Then can we say $f=g$?



      Here nothing is given about the multiplicity $f^{-1}(c_{i})$. So any hint please..



      Thank you.






      complex-analysis polynomials







      share|cite|improve this question











      $endgroup$




      Let $f(x)$ and $g(x)$ are two complex polynomials such that $f^{-1}(c_{i})=g^{-1}(c_{i})$ for two distinct complex numbers $c_{i}$, $i=1,2$.
      Then can we say $f=g$?



      Here nothing is given about the multiplicity $f^{-1}(c_{i})$. So any hint please..



      Thank you.






      complex-analysis polynomials




      complex-analysis polynomials






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 9 '18 at 11:28









      Martin R

      28.8k33356




      28.8k33356










      asked Dec 8 '18 at 17:32









      JOHNJOHN

      485




      485






















          1 Answer
          1






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          1












          $begingroup$

          Either $f$ and $g$ are both constant, or $f=g$. For a proof, assume that
          (without loss of generality), $deg f ge deg g$, and consider the function
          $$
          h(z) = frac{f'(z)(f(z) - g(z))}{(f(z)-c_1)(f(z) - c_2)} , .
          $$
          Show that





          • $h$ has only removable singularities, and therefore can be extended
            to an entire function.


          • $h$ is bounded, and therefore constant.


          • $h$ is identically zero.


          Or, without using complex analysis: Every zero of $(f-c_1)(f-c_2)$ is a zero of $f'(f-g)$ with at least the same multiplicity, therefore
          $$
          f'(f-g) = h(f-c_1)(f-c_2)
          $$
          for some polynomial $h$. Now compare the degrees to conclude that $h = 0$.




          This does not hold for entire functions in general. As an example,
          $f(z) = e^z$ and $g(z) = e^{-z}$ have the same preimage for $c_1 = 0$, $c_2 = 1$, and $c_3 = -1$.



          Rolf Nevanlinna showed 1929 that two (in $Bbb C$) meromorphic functions are identical if they have the same preimages for five distinct values, that is the so-called “Five-Value Theorem.”






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks. Here , to show that h is bounded, we need f and g to be polynomial. But is it true if f and g are just holomorphic functions ?
            $endgroup$
            – JOHN
            Dec 8 '18 at 18:46






          • 1




            $begingroup$
            @JOHN: No – see update.
            $endgroup$
            – Martin R
            Dec 8 '18 at 19:14











          Your Answer





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          1 Answer
          1






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          Either $f$ and $g$ are both constant, or $f=g$. For a proof, assume that
          (without loss of generality), $deg f ge deg g$, and consider the function
          $$
          h(z) = frac{f'(z)(f(z) - g(z))}{(f(z)-c_1)(f(z) - c_2)} , .
          $$
          Show that





          • $h$ has only removable singularities, and therefore can be extended
            to an entire function.


          • $h$ is bounded, and therefore constant.


          • $h$ is identically zero.


          Or, without using complex analysis: Every zero of $(f-c_1)(f-c_2)$ is a zero of $f'(f-g)$ with at least the same multiplicity, therefore
          $$
          f'(f-g) = h(f-c_1)(f-c_2)
          $$
          for some polynomial $h$. Now compare the degrees to conclude that $h = 0$.




          This does not hold for entire functions in general. As an example,
          $f(z) = e^z$ and $g(z) = e^{-z}$ have the same preimage for $c_1 = 0$, $c_2 = 1$, and $c_3 = -1$.



          Rolf Nevanlinna showed 1929 that two (in $Bbb C$) meromorphic functions are identical if they have the same preimages for five distinct values, that is the so-called “Five-Value Theorem.”






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks. Here , to show that h is bounded, we need f and g to be polynomial. But is it true if f and g are just holomorphic functions ?
            $endgroup$
            – JOHN
            Dec 8 '18 at 18:46






          • 1




            $begingroup$
            @JOHN: No – see update.
            $endgroup$
            – Martin R
            Dec 8 '18 at 19:14
















          1












          $begingroup$

          Either $f$ and $g$ are both constant, or $f=g$. For a proof, assume that
          (without loss of generality), $deg f ge deg g$, and consider the function
          $$
          h(z) = frac{f'(z)(f(z) - g(z))}{(f(z)-c_1)(f(z) - c_2)} , .
          $$
          Show that





          • $h$ has only removable singularities, and therefore can be extended
            to an entire function.


          • $h$ is bounded, and therefore constant.


          • $h$ is identically zero.


          Or, without using complex analysis: Every zero of $(f-c_1)(f-c_2)$ is a zero of $f'(f-g)$ with at least the same multiplicity, therefore
          $$
          f'(f-g) = h(f-c_1)(f-c_2)
          $$
          for some polynomial $h$. Now compare the degrees to conclude that $h = 0$.




          This does not hold for entire functions in general. As an example,
          $f(z) = e^z$ and $g(z) = e^{-z}$ have the same preimage for $c_1 = 0$, $c_2 = 1$, and $c_3 = -1$.



          Rolf Nevanlinna showed 1929 that two (in $Bbb C$) meromorphic functions are identical if they have the same preimages for five distinct values, that is the so-called “Five-Value Theorem.”






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks. Here , to show that h is bounded, we need f and g to be polynomial. But is it true if f and g are just holomorphic functions ?
            $endgroup$
            – JOHN
            Dec 8 '18 at 18:46






          • 1




            $begingroup$
            @JOHN: No – see update.
            $endgroup$
            – Martin R
            Dec 8 '18 at 19:14














          1












          1








          1





          $begingroup$

          Either $f$ and $g$ are both constant, or $f=g$. For a proof, assume that
          (without loss of generality), $deg f ge deg g$, and consider the function
          $$
          h(z) = frac{f'(z)(f(z) - g(z))}{(f(z)-c_1)(f(z) - c_2)} , .
          $$
          Show that





          • $h$ has only removable singularities, and therefore can be extended
            to an entire function.


          • $h$ is bounded, and therefore constant.


          • $h$ is identically zero.


          Or, without using complex analysis: Every zero of $(f-c_1)(f-c_2)$ is a zero of $f'(f-g)$ with at least the same multiplicity, therefore
          $$
          f'(f-g) = h(f-c_1)(f-c_2)
          $$
          for some polynomial $h$. Now compare the degrees to conclude that $h = 0$.




          This does not hold for entire functions in general. As an example,
          $f(z) = e^z$ and $g(z) = e^{-z}$ have the same preimage for $c_1 = 0$, $c_2 = 1$, and $c_3 = -1$.



          Rolf Nevanlinna showed 1929 that two (in $Bbb C$) meromorphic functions are identical if they have the same preimages for five distinct values, that is the so-called “Five-Value Theorem.”






          share|cite|improve this answer











          $endgroup$



          Either $f$ and $g$ are both constant, or $f=g$. For a proof, assume that
          (without loss of generality), $deg f ge deg g$, and consider the function
          $$
          h(z) = frac{f'(z)(f(z) - g(z))}{(f(z)-c_1)(f(z) - c_2)} , .
          $$
          Show that





          • $h$ has only removable singularities, and therefore can be extended
            to an entire function.


          • $h$ is bounded, and therefore constant.


          • $h$ is identically zero.


          Or, without using complex analysis: Every zero of $(f-c_1)(f-c_2)$ is a zero of $f'(f-g)$ with at least the same multiplicity, therefore
          $$
          f'(f-g) = h(f-c_1)(f-c_2)
          $$
          for some polynomial $h$. Now compare the degrees to conclude that $h = 0$.




          This does not hold for entire functions in general. As an example,
          $f(z) = e^z$ and $g(z) = e^{-z}$ have the same preimage for $c_1 = 0$, $c_2 = 1$, and $c_3 = -1$.



          Rolf Nevanlinna showed 1929 that two (in $Bbb C$) meromorphic functions are identical if they have the same preimages for five distinct values, that is the so-called “Five-Value Theorem.”







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 10 '18 at 8:59

























          answered Dec 8 '18 at 17:51









          Martin RMartin R

          28.8k33356




          28.8k33356












          • $begingroup$
            Thanks. Here , to show that h is bounded, we need f and g to be polynomial. But is it true if f and g are just holomorphic functions ?
            $endgroup$
            – JOHN
            Dec 8 '18 at 18:46






          • 1




            $begingroup$
            @JOHN: No – see update.
            $endgroup$
            – Martin R
            Dec 8 '18 at 19:14


















          • $begingroup$
            Thanks. Here , to show that h is bounded, we need f and g to be polynomial. But is it true if f and g are just holomorphic functions ?
            $endgroup$
            – JOHN
            Dec 8 '18 at 18:46






          • 1




            $begingroup$
            @JOHN: No – see update.
            $endgroup$
            – Martin R
            Dec 8 '18 at 19:14
















          $begingroup$
          Thanks. Here , to show that h is bounded, we need f and g to be polynomial. But is it true if f and g are just holomorphic functions ?
          $endgroup$
          – JOHN
          Dec 8 '18 at 18:46




          $begingroup$
          Thanks. Here , to show that h is bounded, we need f and g to be polynomial. But is it true if f and g are just holomorphic functions ?
          $endgroup$
          – JOHN
          Dec 8 '18 at 18:46




          1




          1




          $begingroup$
          @JOHN: No – see update.
          $endgroup$
          – Martin R
          Dec 8 '18 at 19:14




          $begingroup$
          @JOHN: No – see update.
          $endgroup$
          – Martin R
          Dec 8 '18 at 19:14


















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