Correct this problem in complex numbers

Question

Prove for all $|z| = 3$, $$\frac{8}{11} \leq \left | \frac{z^2 + 1}{z^2 + 2}\right | \leq \frac{10}{7}.$$

Here is what I did,

$$\frac{8}{11}\leq \frac{8}{z^2 + 2}=\frac{|z^2| - 1}{z^2 +2}\leq\left | \frac{z^2 + 1}{z^2 + 2} \right |= \left | \frac{z^2}{z^2+ 2} + \frac{1}{z^2+2} \right | \leq \left | \frac{z^2}{z^2+ 2} \right | + \left | \frac{1}{z^2+ 2} \right | = \frac{|z^2|}{z^2 + 2} + \frac{1}{z^2 + 2} = \frac{10}{z^2 + 2 } \leq \frac{10}{9} \leq \frac{10}{7}$$

If i misused any properties, typo or not, can you identify them for me?

Answer 1

$8=|z^2|-1\leq |z^2+1|\leq |z^2|+1=10$ and $7=|z^2|-|2|\leq |z^2+2|\leq |z^2|+2=11$

Answer 2

You should not have expression involving $z$ without absolute value signs. For example, the statement $$\frac{8}{11} \leq \frac{8}{z^2+2}$$ makes no sense as $\mathbb{C}$ is not an ordered field, while $$\frac{8}{11} \leq \frac{8}{|z|^2+2}$$ does make sense (and is true), as $\mathbb{R}$ is an ordered field.

Taking this into consideration, you have successfully shown $$\frac{8}{11} \leq \left|\frac{z^2 + 1}{z^2 + 2}\right| \leq \frac{10}{|z^2 + 2|}.$$ You then claim $$\frac{10}{|z^2 + 2|} \leq \frac{10}{9}.$$ I don't see how you obtained this. Instead, you should use the reverse triangle inequality on $|z^2 + 2|$ to obtain the desired upper bound for the given expression.

Answer 3

You can establish a slighly stronger inequality by noticing that $|z|=3$ implies $|z^2+2|\in[7,11]$, hence: $$\left|\frac{z^2+1}{z^2+2}\right|=\left|1-\frac{1}{z^2+2}\right|\in\left[\frac{6}{7},\frac{12}{11}\right].$$

Answer 4

Like For all complex numbers with $|z|=2$, inequality $2\le |z-4|\le 6$ holds,

$\displaystyle z=3(\cos\theta+i\sin\theta)\implies z^2=9(\cos2\theta+i\sin2\theta)$

$\displaystyle\left|\frac{z^2+1}{z^2+2}\right|=\sqrt{\frac{82+18\cos2\theta}{85+36\cos2\theta}}$

Now clearly, the maximum and minimum values of $\displaystyle\frac{82+18\cos2\theta}{85+36\cos2\theta}=\frac12+\frac{79}{2(85+36\cos2\theta)}$ are attainable at $\cos2\theta=-1,1$ respectively which are $\displaystyle\frac{82+18(-1)}{85+36(-1)}=\left(\frac87\right)^2$ and $\displaystyle\frac{82+18(+1)}{85+36(+1)}=\left(\frac{10}{11}\right)^2$ respectively