Trying prove that the complex function is bounded

Question

I am trying to prove that, for fixed $\theta \in (\frac{\pi}{2},\pi)$ we can to bound the function $f:(\epsilon, \infty) \to \mathbb{K}$, $\epsilon>0$ $$f(s)=\frac{s e^{i2\theta}+e^{i\theta}}{s^2 e^{i3\theta}+e^{i\theta}}$$ by a number $N_{\theta}$ (for example) or a function $s$-dependent. If $s>1$, we can do it by absolute value properties, but in general i didn't can do it.

Answer 1

It can be bounded by $N_\theta$ (dependent on $\theta$), but not uniformly in $\theta$.

I would write $z = e^{i\theta}$. Note that $z$ lies on the unit circle, in the second quadrant. We have \begin{align*} f(s) & = \frac{sz^2 + z}{s^2 z^3 + z}\\ & = \frac{sz + 1}{s^2 z^2 + 1}\\ & = \frac{w + 1}{w^2 + 1}\\ & = \frac{w + 1}{(w + i) (w - i)}\\ & = \frac{A}{w + i} + \frac{B}{w - i} \end{align*} where $w = sz$, $A = \frac{i - 1}{2i}$, $B = \frac{i + 1}{2i}$. For a fixed $\theta \in (\pi/2, \pi)$, as $s$ goes over $(\epsilon, \infty)$, $w = se^{i\theta}$ sweeps over a ray in the second quadrant, originating from $\epsilon e^{i\theta}$, going to $\infty$. Observe that for $w$ along this ray, we have $$ |w - i| \geq \sin(\theta - \pi/2), $$ and $$ |w + i| \geq 1. $$ So we have the bound $$ |f(s)| \leq |A| + \frac{|B|}{\sin(\theta - \pi/2)}. $$

We can't bound $f(s)$ uniformly in $\theta$ though. When $s = 1$, as $\theta \rightarrow \pi/2$, $w = se^{i\theta}$ approaches $i$, and the term $\frac{B}{w - i}$ blows up.

A bound in terms of $s$ is also possible, provided $s \neq 1$. In this case, $w = se^{i\theta}$ lies on portion of the circle of radius $s$ in the second quadrant. As long as it stays away from $\pm i$, we can bound $f(s)$.