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?
$\left|\frac{z}{4}+\frac{6}{z}\right|=\left|\frac{z^2+24}{4z}\right|$

The function $f(z)=\frac{z^2+24}{4z}$ is analitic on $1\leq|z|\leq7$. Therefore, by the maximum modulus theorem its maximum absolute value is attained at the boundary. The boundary are the circles $|z|=1$ and $|z|=7$.

For $|z|=1$, observe that $z^2$ just travels the same circle. We have $|f(z)|=|z^2+24|/4$, which is maximum for $z=1$ or $z=-1$ (such that $z^2$ and $24$ point in the same direction).

For $|z|=7$, observe that $z^2$ travels the circle $|w|=49$. We have $|f(z)|=|z^2+24|/28$, which is maximum for $z=7$ or $z=-7$ (such that $z^2$ and 24 point in the same direction).

So, $f(7)=f(-7)$ seem to be the largest.

The minimum is zero at $f(\pm\sqrt{24}i)$.
We want to find the minimum and maximum of 

$\left| \dfrac{z}{4}+\dfrac{6}{z}\right|$

When $1\le |z|\le 7$

Let $z=x+iy$ and plug it in the given expression

$$\left| \dfrac{x+iy}{4}+\dfrac{6}{x+iy}\right|=\left| \dfrac{x+iy}{4}+\dfrac{6(x-iy}{x^2+y^2}\right|=\left|\frac{6 x}{x^2+y^2}+\frac{x}{4}+i\left(\frac{y}{4}-\frac{6 y}{x^2+y^2}\right)\right|$$

Which is 
$$r=\sqrt{\left(\frac{6 x}{x^2+y^2}+\frac{x}{4}\right)^2+\left(\frac{y}{4}-\frac{6 y}{x^2+y^2}\right)^2}=\frac{\sqrt{r^4-48 r^2+96 x^2+576}}{4 r}$$
where $r^2=x^2+y^2$

As we have $1\le r\le 7$ let's plug $r=1$ and $r=7$ in the previous expression

for $r=1$ we get $\dfrac{ \sqrt{96 x^2+529}}{4}$ which is $\dfrac{25}{4}$ for $x=1$, maximum

for $r=7$ we get $\dfrac{\sqrt{96 x^2+625}}{28} $ which is $\dfrac{73}{28}$ for $x=7$, minimum

Hope this helps 

Edit 

Actually $\dfrac{z}{4}+\dfrac{6}{z}=0$ has a solution at $z=\pm 2 i \sqrt{6}$ whose module satisfies the condition $|z|\in[1,\;7]$ therefore that is the absolute minimum, zero.