Assume that an entire function $f$ be finite order with finitely many zeros.

Please show that either $f(z)$ is a polynomial or $f(z) + z$ has infinitely many zeros.

Thank you.

And I know the following theorem,

Suppose f is entire function of finite order. Then either f has infinitely many zeros or $f(z)$ is of the form $Q(z)e^{P(z)}$ for polynomials $P$ and $Q$.

  • $\begingroup$ Hint: Does $f(z)+z$ have finite order? Does $f(z)+z$ have the form $Q(z)e^{P(z)}$? $\endgroup$ – Antonio Vargas Dec 22 '13 at 20:37
  • $\begingroup$ Can you write this more explicitly? By assuming that $f(z)+z$ has finitely many zeros, I'll get a condtradiction. Should I prove so? If so, can you show this proof? Thank you. @AntonioVargas $\endgroup$ – user315 Dec 22 '13 at 21:33

Finite order is not needed. Apply Picard's "big" theorem to $f(z)/z$.

  • 1
    $\begingroup$ $f(z)/z$ has an essential singularity at $\infty$. The theorem then says that with at most one exception, $f(z)/z$ takes every complex value infinitely many times. But by assumption $0$ is that exception, so $-1$ is not... $\endgroup$ – Robert Israel Dec 22 '13 at 21:05
  • $\begingroup$ Why $f(z)/z$ has an essential singularity at $\infty$ and not a pole ? $\endgroup$ – WLOG Dec 22 '13 at 21:12
  • 1
    $\begingroup$ @boris: A meromorphic function on the Riemann sphere is a rational function. $\endgroup$ – Robert Israel Dec 22 '13 at 23:49

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