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in the following problem I have some questions.

Show that if $f$ is an entire function of finite order thar omits two values, then $f$ is constant.

I know that by Hadamard products, if $f$ omits the values $a$ and $b$, I can write $f(z)-a=e^{p(z)}$ and $f(z)-b=e^{q(z)}$, where $p(z)$, $q(z)$ are polynomials of degree $n,m$ respectively. Then $$e^{p(z)}-e^{q(z)}=b-a.$$

Now, if $z\to\infty$ the leading terms of $p$ and $q$ must be the same, why? And then, how can I conclude that $f$ is constant?

Thanks in advance !

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We suppose that $n\geq 1$ and $m\geq 1$. From $\exp(p)-\exp(q)=b-a$ with $b\not = a$, we deduce that $p^{\prime}\exp(p)-q^{\prime}\exp(q)=0$, and that the two polynomials $p^{\prime}$ and $q^{\prime}$ have the same zeros with the same multiplicities. Hence there exists a constant $c$ such that $q^{\prime}=c p^{\prime}$. Hence we get that $\exp(p)=c\exp(q)$, and that $g=(c-1)\exp(q)=b-a$, (hence $c\not =1$). Now compute the derivative of $g$ to finish the job.

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  • $\begingroup$ Works fine, thank you :) $\endgroup$ – jnaf Jun 8 '17 at 2:04

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