show that $\frac{R^4-R}{R^2+R+1}\le |\frac{z^4+iz}{z^2+z+1}| \le \frac{R^4+R}{(R-1)^2}$

Question

show that $\frac{R^4-R}{R^2+R+1}\le |\frac{z^4+iz}{z^2+z+1}| \le \frac{R^4+R}{(R-1)^2}$

for all $z$ satisfying $|z|=R>1$

My idea:

$|\frac{z^4+iz}{z^2+z+1}| \le |\frac{|z^4|+|iz|}{|z^2|+|z|+1}|\le \frac{R^4+R}{R^2+R+1} $

Answer 1

For the denominator: $|z^{2}+z+1|\geq|z|^{2}-|z|-1=R^{2}-R-1\geq R_{2}-2R-1=(R-1)^{2}$, so $\dfrac{1}{|z^{2}+z+1|}\leq\dfrac{1}{(R-1)^{2}}$.

For the denominator again: $|z^{2}+z+1|\leq|z|^{2}+|z|+1=R^{2}+R+1$, so $\dfrac{1}{|z^{2}+z+1|}\geq\dfrac{1}{R^{2}+R+1}$.

For the numerator: $|z^{4}+iz|\geq|z|^{4}-|iz|=R^{4}-R$.