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.


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.”

  • $\begingroup$ 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

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