Computing an upper bound in the complex plane

Question

While working through the basics of compelx analysis, I stumbled upon this probelm whereby I need to find the upper bound of:

$\lvert\frac{z^2-4z-2}{z^3+2z^2-4}\rvert$$ \forall z \in \mathbb{C}$ s.t $\lvert{z}\rvert = 3$

I'm quite stumped but I assume that I need the denominator to be as small as possible and the numerator to be as big as possible, hence using the triangle inequality and the reverse triangle inequality to solve this. Can anyone start me off?

Much appreciated

Answer 1

$|z^{3}+2z^{2}-4|\geq|z|^{3}-2|z|^{2}-4=3^{3}-2\cdot 3^{2}-4=5$, and $|z^{2}-4z-2|\leq|z|^{2}+4|z|+2=3^{2}+4\cdot 3+2=23$, so an upper bound can be taken as $23/5$.

Answer 2

Just for your information: The maximum absolute value of your expression is about $1.55. See the following output: