Q:

How do I see that $|Re^{it}((Re^{it})^2+4)| \ge R(R^2 - 4)$ for $R \in \mathbb R$ sufficiently large?

How do I see that $|Re^{it}((Re^{it})^2+4)| \ge R(R^2 - 4)$ for  $R \in \mathbb R$ sufficiently large ?
I've been trying to use triangle inequality and other ways, but I've not come to a conclusion.
Can someone help me out ?

A:

Just rewrite it using Euler's formula:
$$Re^{it}(4 + R^2 e^{2it}) = R^3e^{3it}+4Re^{it} = (R^3\cos(3t) + 4R\cos(t)) + i(R^3\sin(3t) + 4R\sin(t)).$$
Hence
$$|Re^{it}(4 + R^2 e^{2it})|^2 = (R^3\cos(3t) + 4R\cos(t))^2 + (R^3\sin(3t) + 4R\sin(t))^2;$$
simplify it and
$$|Re^{it}(4 + R^2 e^{2it})|^2 = R^2 (R^4 - 8R^2+16 + 16R^2\cos^2(t)) \ge R^2 (R^4 - 8R^2+16) = R^2(R^2-4)^2;$$
so you have
$$|Re^{it}(4 + R^2 e^{2it})| \ge R(R^2 - 4)$$ for $R\ge 2$.

