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$$\frac{R^{4}-R}{R^{2}+R+1} \leq\left|\frac{z^{4}+i z}{z^{2}+z+1}\right| \leq \frac{R^{4}+R}{(R-1)^{2}}$$
for all $z$ satisfying $|z|=R>1$.
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Proof. When $|z|=R>1$,
$$\left|z^{4}+i z\right| \geq\left|z^{4}\right|-|i z|=|z|^{4}-|i||z|=R^{4}-R$$
and
$$\left|z^{2}+z+1\right| \leq\left|z^{2}\right|+|z|+|1|=|z|^{2}+|z|+1=R^{2}+R+1$$
by triangle inequality. Hence
$$\left|\frac{z^{4}+i z}{z^{2}+z+1}\right| \geq \frac{R^{4}-R}{R^{2}+R+1}$$
On the other hand,
$$\left|z^{4}+i z\right| \leq\left|z^{4}\right|+|i z|=|z|^{4}+|i||z|=R^{4}+R$$

and
\begin{aligned} \left|z^{2}+z+1\right| &=\left|\left(z-\frac{-1+\sqrt{3} i}{2}\right)\left(z-\frac{-1-\sqrt{3} i}{2}\right)\right| \\ &=\left|z-\frac{-1+\sqrt{3} i}{2}\right|\left|z-\frac{-1-\sqrt{3} i}{2}\right| \\ & \geq\left(|z|-\left|\frac{-1+\sqrt{3} i}{2}\right|\right)\left(|z|-\left|\frac{-1-\sqrt{3} i}{2}\right|\right) \\ &=(R-1)(R-1)=(R-1)^{2} \end{aligned}
Therefore,
$$\left|\frac{z^{4}+i z}{z^{2}+z+1}\right| \leq \frac{R^{4}+R}{(R-1)^{2}}$$
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