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?
 A: $\left|\frac{z}{4}+\frac{6}{z}\right|=\left|\frac{z^2+24}{4z}\right|$
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 $f(\pm\sqrt{24}i)$.
A: $|\frac {z}{4} + \frac {6}{z}| \ge 0$
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 $|\frac {z}{4}| + |\frac {6}{z}|$
A: 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 $\sqrt{36.0625+3}=6.25$.
A: 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.
