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Let $ f $ a complex entire function such that:

$ |f(z)| \leq \sqrt{2|z|} + \frac{1}{\sqrt{2|z|}} \quad \forall z \neq 0 $

Prove that $ f$ is constant.

Thank's in advance!

2 Answers 2

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Let $g(z)=\frac{f(z)-f(0)}{z}$. Then $g(z)$ is entire and for all $z$ with $|z| \geq 1$ you have

$ \left| g(z) \right| \leq \left|\frac{f(z)}{z} \right| + \left|\frac{f(0)}{z} \right| \leq \left|\frac{\sqrt{2|z|} + \frac{1}{\sqrt{2|z|}}}{z} \right| + \left|f(0) \right| $

Now from here it is easily to prove that $g$ is bounded (note that by continuity $g$ is bounded on $|z| \leq 1$), thus, since entire is constant.

This proves that $f(z)=az+b $ for some $a,b$. Plug this in your equation and see that $a =0$.

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    O.K. Thank you!2012-07-05
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Since $f(z)$ is entire, from Cauchy integral formula, we have $f^{(n)}(0) = \dfrac{n!}{2 \pi i}\oint_{C_r} \dfrac{f(z)}{z^{n+1}} dz$ Now argue out why $f^{(n)}(0) = 0$ for $n > 0$ by letting $r \to \infty$ and looking at what happens to the integral.

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    O.K Now is clear! Thank you very much for your time!2012-07-05