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

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.”

• 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 ? – JOHN Dec 8 '18 at 18:46
• @JOHN: No – see update. – Martin R Dec 8 '18 at 19:14