I need to show that he upper bound of $\frac{z^{3/2}}{1 + z^3} = 0$ when $R \to \infty$, where $z = R(cos\theta +i sin\theta)$
I'm trying to use the estimation lemma, but the power of 3 confuses me.
$$|\frac{z^{3/2}}{1 + z^3}| = |\frac{R^{3/2}e^{3i\theta/2}}{1 + R^3e^{i3\theta}}| $$
because of the complex conjugate
$$\frac{|R^{3/2}e^{3i\theta/2}|}{|1 + R^3e^{i3\theta}|} = \frac{R^{3/2}}{|1 + R^3e^{i3\theta}|} $$
However, since I have $R^3$, I can't just take it out of the absolute value.
Any help will be appreciate.