<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] Let $x = r^2 \implies 1 \le x \le 49 $, and consider $f(x) = \dfrac{x}{16} + \dfrac{36}{x}\implies f'(x) = \dfrac{1}{16}- \dfrac{36}{x^2}\implies f'(x) = 0 \iff x = 24$, and $f(1) = 36+\dfrac{1}{16} = 36.0625, f(49) = \dfrac{49}{16}+\dfrac{36}{49}=3.8, f(24) = 3.$ $3\cos(2\theta)$ has a min of $-3$, and a max of $3$. Put these values together, we have: $\left|\dfrac{z}{4}+\dfrac{6}{z}\right|$ reaches a min of $\sqrt{3.8-3}\approx. 0.89$, and a max of $\sqrt{36.0625+3}=6.25$.
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