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$\lim_{z \to \infty} \frac{\operatorname{Re} f(z)}{z}=0 $ implies $f$ is constant.

$f$ is entire function on complex plane

I tried to bound some derivative of $e^{f(z)}$ by cauchy integral inequality hence $e^{f(z)}$ is polynomial.

But since it has exponential term depending on radius, and denominator has $R^n$ for radius $R$ so i can not bound it.

Should I try some other way?

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    what is $f$? ${}{}{}$ – 2017-01-07
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    By $Ref(z)$ do you mean $\Re(f(z))$ (the real part), or $Re^{f(z)}$? I'd normally intepret it the first way, but you mention both $R$ and $e^{f(z)}$ later in your post. – 2017-01-07
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    oh sorry, f is entire function – 2017-01-07
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    @mark it Re f(z) means real part of f – 2017-01-07
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    not a duplicate of [Hints for a complex limit...](http://math.stackexchange.com/questions/528517/hints-for-a-complex-limit-prove-if-lim-z-to-infty-fz-z-0-then-fz). $\lim_{z\to \infty} f(z)/z=0 \implies \lim_{z\to \infty} (\operatorname{Re}f(z))/z=0$ but $\lim_{z\to \infty} (\operatorname{Re}f(z))/z=0$ does not imply $ \lim_{z\to \infty} f(z)/z=0.$ – 2017-01-07

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