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.

$\endgroup$
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.”

$\endgroup$
  • $\begingroup$ 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

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.