<s>[INST] Given $1 \le |z| \le 7$ Find least and Greatest values of $\left|\frac{z}{4}+\frac{6}{z}\right|$ Given $1 \le |z| \le 7$
Find least and Greatest values of $\left|\frac{z}{4}+\frac{6}{z}\right|$
I have taken $z=r e^{i \theta}$ $\implies$ $1 \le r \le 7$
Now $$\left|\frac{z}{4}+\frac{6}{z}\right|=\left|\frac{r \cos \theta}{4}+\frac{ir \sin \theta}{4}+\frac{6 \cos \theta}{r}-\frac{6 i \sin \theta}{r} \right|$$
So
$$\left|\frac{z}{4}+\frac{6}{z}\right|=\sqrt{\frac{r^2}{16}+\frac{36}{r^2}+3 \cos (2\theta)}$$
any clue from here?
 [/INST] $|\frac {z}{4} + \frac {6}{z}| \ge 0$
I say that there exist a $z$ such that $1\le|z|\le 7$ and $\frac {z}{4} + \frac {6}{z} = 0$
and any such $z$ must minimize the objective.
$z = i 2\sqrt {6} $
To maximize the objective
$|\frac {z}{4} + \frac {6}{z}| \le |\frac {z}{4}| + |\frac {6}{z}|$
if $z$ is real then:
$|\frac {z}{4} + \frac {6}{z}| = |\frac {z}{4}| + |\frac {6}{z}|$
$z = 7$ maximises $|\frac {z}{4}| + |\frac {6}{z}|$
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