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Let $f: \mathbb{C} \rightarrow \mathbb{C}$ be a non-constant entire function with the property that $f(x) \in \mathbb{R}$ if $x \in \mathbb{R}.$

Then I would like to know if it already follows that for any $z_0 \in \mathbb{R}$ there is $R>0$, $C>0$ small enough and $n \in \mathbb{N}$ such that $$C \left\lvert {\bf Im} (z^n) \right\rvert \le \left\lvert {\bf Im} f(z) \right\rvert$$

for any $z \in B_{\mathbb{C}}(z_0,R)$ where $z_0 \in \mathbb{R}.$

${\bf Im}$ denotes imaginary part here.

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    Here's a hint: $n$ should be the exponent in the first nonvanishing term in the power series expansion of $f(z)$ centered at $z_0$. What do you know about all the coefficients in that series? – 2017-02-09

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I think this goes wrong whenever $f'(z_0)=0$ and $z_0\neq 0$.

For example, $f(z)=(z-1)^2$, $z_0=1$. The function of a real variable $g(t)=\mathrm{Im}(f(1+it))$ is identically $0$, whereas $h(t) = \mathrm{Im}((1+it)^n)$ is strictly increasing in small neighborhoods of $0$, so the inequality cannot hold.

To generalize, if $f'(z_0)=0$, then $g(t)=\mathrm{Im}f(z_0+it)$ will have a horizontal tangent line at $0$, whereas $h(t)=\mathrm{Im}((z_0+it)^n)$ has slope $nz_0^{n-1}$ at $0$, so if $z_0\neq 0$ you can get for all $C>0$, $t$ arbitrarily small with $|g(t)|<\dfrac{Cnz_0^{n-1}}2t$ while $|h(t)|>\dfrac{nz_0^{n-1}}{2}t$.