The greatest value of $|z|$ if $\Big|z+\frac{1}{z}\Big|=3$ where $z\in\mathbb{C}$

Question

$\bigg|z+\dfrac{1}{z}\bigg|=3$ then the greatest value of $|z|$ is ___________

My Attempt $$ \bigg|z+\frac{1}{z}\bigg|=\bigg|\dfrac{z^2+1}{z}\bigg|=\frac{|z^2+1|}{|z|}=3\\ \bigg|z+\frac{1}{z}\bigg|=3\leq|z|+\frac{1}{|z|}\implies |z|^2-3|z|+1\ge0\\ |z|=\frac{3\pm\sqrt{5}}{2}\implies\color{red}{|z|\in(-\infty,\frac{3-\sqrt{5}}{2}]\cup[\frac{3+\sqrt{5}}{2},+\infty)} $$ $$ \bigg|z+\frac{1}{z}\bigg|=3\geq\bigg||z|-\frac{1}{|z|}\bigg|\\ |z|^2-3|z|-1\leq0\text{ or }|z|^2+3|z|-1\geq0\\ |z|=\frac{3\pm\sqrt{13}}{2}\text{ or }|z|=\frac{-3\pm\sqrt{13}}{2}\\ \color{red}{|z|\in[\frac{3-\sqrt{13}}{2},\frac{3+\sqrt{13}}{2}]}\text{ or }\color{red}{|z|\in(-\infty,\frac{-3-\sqrt{13}}{2}]\cup[\frac{-3+\sqrt{13}}{2},+\infty)} $$ The solution given in my reference is $\dfrac{3+\sqrt{13}}{2}$, why am I not able to find it in my attempt ?

Note: A similar question has been asked before, If $∣z+\frac{1}{z}∣=a$ then find the minimum and maximum value of $|z|$, but it does not have any answer which address why we choose a specific version of the triangle inequality.

Answer 1

(First, assume that $|z|>1$; our range for $|z|$ will be mirrored with its reciprocal, but the maximum we're looking for is definitely greater than $1$.)

Your inequality $\left|z+\frac1z\right|\le |z|+\left|\frac1z\right|$ is true, but it's only half the picture; we get equality when $z$ and $\frac1z$ are positive real multiples of each other, which makes $|z|$ as small as possible. Solving the resulting quadratic inequality $|z|^2-3|z|+1 \ge 0$, we get $|z|\ge\frac{3+\sqrt{5}}{2}$ or $|z|\le \frac{3-\sqrt{5}}{2}$.

We should also look at an inequality in the other direction: $\left|z+\frac1z\right|\ge |z|-\left|\frac1z\right|$. This leads to the quadratic inequality $|z|^2-3|z|-1\le 0$, with solutions $\frac{3-\sqrt{13}}{2}\le z\le \frac{3+\sqrt{13}}{2}$.

Combine the two inequalities and our restriction $|z|>1$, and we get the range $$\frac{3+\sqrt{5}}{2} \le |z|\le \frac{3+\sqrt{13}}{2}$$ with equality achieved at pure real values $\pm\frac{3+\sqrt{5}}{2}$ for the lower bound and pure imaginary values $\pm i\frac{3+\sqrt{13}}{2}$ for the upper bound.

So, then, your work - you solved the quadratic wrong, and got the numbers from the other quadratic. (Edit note - fixed now) Then your reference had to go mix things up with an answer that's just plain wrong. No wonder you're confused here.

...why we choose a specific version of the triangle inequality.

The real answer? We choose all of them. They all apply, restricting different aspects of things. Some of those restrictions won't matter for the specific problem we want to solve, but they're all there.

(In response to the red text added)
There's a reason I added the assumption that $|z|>1$; your solution to the inequality $3\ge \left||z|-\frac1{|z|}\right|$ mixes two cases together into a useless mush. Combine the two parts of what you wrote, and it's no restriction on $|z|$ at all - all positive values are in at least one of those ranges.

If $|z|\ge 1$, then $|z|\ge\frac1{|z|}$ and we can write $\left||z|-\frac1{|z|}\right|=|z|-\frac1{|z|}$. This leads to the inequality $|z|^2-3|z|-1\le 0$ and solutions $\frac{3-\sqrt{13}}{2}\le z\le \frac{3+\sqrt{13}}{2}$. But then, we also assumed $|z|\ge 1$, so that's really $1\le z\le \frac{3+\sqrt{13}}{2}$.

If $|z|\le 1$, then $|z|\le\frac1{|z|}$ and we can write $\left||z|-\frac1{|z|}\right|=\frac1{|z|}-|z|$. This leads to the inequality $|z|^2+3|z|-1\ge 0$, with solutions $|z|\le\frac{-3-\sqrt{13}}{2}$ or $|z|\ge \frac{-3+\sqrt{13}}{2}$. The first of those? We can't have $|z|$ negative. Bye. The second? Combine with the case's condition for $\frac{\sqrt{13}-3}{2}\le |z|\le 1$.

So then, the full implication of that inequality is $$\frac{2}{3+\sqrt{13}}=\frac{\sqrt{13}-3}{2}\le |z|\le\frac{3+\sqrt{13}}{2}$$ Combine with the other half, and we get two ranges for $z$: $$\frac{2}{3+\sqrt{13}}=\frac{\sqrt{13}-3}{2}\le |z|\le \frac{3-\sqrt{5}}{2}=\frac{2}{3+\sqrt{5}}\quad\text{or}\quad\frac{3+\sqrt{5}}{2}\le |z|\le \frac{3+\sqrt{13}}{2}$$

Answer 2

The Joukowsky transform $z\to z+1/z$ transforms circles of radius $r>1$ into ellipses of semi-axes $r-1/r$ and $r+1/r$. The max value of $|z|$ corresponds to the radius of the circle such that the smallest semi-axis $r-1/r=3$. Solving this equation $r^2-3r-1=0$ for $r>0$ you get the result $r=(3+\sqrt{13})/2$ which agrees with yours.

The minor semi axes correspond to the purely imaginary values that jmerry found.

Answer 3

A possible approach is using polar representation

• $z = |z|e^{i\phi}$
• $\Big|z+\frac{1}{z}\Big|=3 \Leftrightarrow (z+\frac{1}{z})(\bar z + \frac{1}{\bar z}) = 9 \Leftrightarrow \boxed{\left(|z|e^{i\phi} + \frac{1}{|z|} e^{-i\phi}\right)\left(|z|e^{-i\phi} + \frac{1}{|z|} e^{i\phi}\right) = 9}$

Expanding gives \begin{eqnarray*} |z|^2 + e^{-2i\phi} + e^{2i\phi} + \frac{1}{|z|^2} & = & 9 \\ & \Leftrightarrow & \\ |z|^2 + 2\cos 2\phi + \frac{1}{|z|^2} & = & 9 \\ \end{eqnarray*}

So, the maximum $|z|$ is reached for $\cos 2\phi = -1$, since the maximum solution of $q+\frac{1}{q}= C$ is $q= \frac{1}{2}(C+\sqrt{C^2-4})$ . Hence, solve

\begin{eqnarray*} |z|^2 + \frac{1}{|z|^2} & = & 11 \\ & \stackrel{|z|\: max.}{\Rightarrow} & \\ |z| & = & \sqrt{\frac{11+\sqrt{117}}{2}}\\ \end{eqnarray*}

Here I show that this is the same as the given solution:

\begin{eqnarray*} |z| & = & \sqrt{\frac{11+\sqrt{117}}{2}}\\ & = & \sqrt{\frac{22+2\cdot 3 \sqrt{13}}{4}}\\ & = & \sqrt{\frac{9 + 13+2\cdot 3 \sqrt{13}}{4}}\\ & = & \sqrt{\frac{(3 +\sqrt{13})^2}{4}}\\ & = & \frac{3 +\sqrt{13}}{2}\\ \end{eqnarray*}