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 $f(\pm\sqrt{24}i)$.
