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which of the following are true ?

a) There exists non-constant entire function which is bounded on real and imaginary axis.

b) There exists entire function such that $f(0)=1$ and such that $|f(z)|$ is less than or equal to $\dfrac{1}{|z|}$, for all $|z|$ greater than or equal to $5$.

I think for a) $e^{iz^2}$ works. But I have no idea about (b) can I apply Liouville's theorm for (b) or picard's theorm.?

Function is bounded on and outside the disc $|z|=5$. To apply Liouville's theorm we have to prove function is bounded inside $|z|=5$. After this function becomes constant And $f (z) =1$ then I can contradict.

Thanks in advance.

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  • $\begingroup$ Do they need to be analytic or meromorphic or something like that? Otherwise if no extra criterion it's probably a bit easier. $\endgroup$ – mathreadler Mar 5 '17 at 10:33
  • $\begingroup$ For ( a) function can be f (z)=e^iz^2 .I have no idea for option( b) $\endgroup$ – Gilll Mar 5 '17 at 10:43
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$b)$ If such an $f$ exists, then on $|z|\geq5$, $$ |f(z)| \leq \frac{1}{|z|} \leq \frac{1}{5} \hspace{4cm} (1) $$ Further, $f$ is a continuous function and $|z|\leq5$ is compact, therefore $f$ attains a maximum on $|z|\leq5$. Thus, we obtain that $f$ is bounded everywhere. By Liouville's theorem, $f$ is constant. As $f(0)=1$, thus $f \equiv 1$. This is a contradiction to $(1)$.

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If $f$ is entire, it is continuous on the closed disk $\{|z|\le5\}$, and hence bounded.

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