# Why must $|z|\gt 1$ be the necessary condition

Question:- If $\left|z+\dfrac{1}{z} \right|=a$ where $z$ is a complex number and $a\gt 0$, find the greatest value of $|z|$.

My solution:- From triangle inequality we have

$$|z|-\left|\dfrac{1}{z}\right|\le\left|z+\dfrac{1}{z} \right|\le|z|+\left|\dfrac{1}{z}\right| \implies |z|-\left|\dfrac{1}{z}\right|\le a\le|z|+\left|\dfrac{1}{z}\right|$$ Now on solving the inequalities separately, we get the following $$$$\tag{1}|z|-\left|\dfrac{1}{z}\right|\le a \implies \dfrac{a-\sqrt{a^2+4}}{2}\le|z|\le\dfrac{a+\sqrt{a^2+4}}{2}$$$$

$$$$\tag{2}|z|+\left|\dfrac{1}{z}\right|\ge a \implies |z| \in \mathbb{R}-\left(\dfrac{a-\sqrt{a^2-4}}{2},\dfrac{a+\sqrt{a^2-4}}{2} \right)$$$$

From $(1)$ and $(2)$, we get $$\boxed{|z|_{max}=\dfrac{a+\sqrt{a^2+4}}{2}}$$

My problem with the question:-

The book from which I am solving tells to take note of the following point for the question.

$|z_1+z_2|\ge |z_1|-|z_2|$ and $|z_1+z_2|\ge |z_2|-|z_1|$. Here we have taken $|z|-\dfrac{1}{|z|}$ since we have to find the greatest value of $|z|$ and hence we take the case $|z| \gt 1$

Now all this does is that make the bound tighter nothing else, so why the need of the specific condition $|z|\gt 1$ and also why, only $|z|-\dfrac{1}{|z|}$ provides the maximum value. From, this what I mean to ask is how can we tell even before solving for $|z|$ that $|z|-\dfrac{1}{|z|}$ provides the maximum value

• One of $|z|, |\frac 1z|$ is greater than 1 and one is less than 1, (or they both equal 1) and without loss of generality you can choose $|z|>1.$ You want $|z| - |\frac1z| > 0$ or there isn't any information there since $a>0.$ Jul 15, 2016 at 18:28

why the need of the specific condition $|z|\gt 1$

$$a=\left|z+\frac 1z\right|\ge|z|-\frac{1}{|z|}\tag1$$

If $0\lt |z|\le 1$, then $-\frac{1}{|z|}\le -1$, so $$|z|-\frac{1}{|z|}\le 1-1=0\tag2$$

From $(1)$, we have $$a=\left|z+\frac 1z\right|\ge |z|-\frac{1}{|z|}=(\text{non-positive})$$ which is true since $a\gt 0$, so in this case the maximum value of $|z|$ is $1$. Now, of course, we are interested in the case when $|z|\gt 1$. (so, I think that the book does not say that $|z|\gt 1$ is the necessary condition, and that the book implies that the case $0\lt |z|\le 1$ is trivial.)

why, only $|z|-\dfrac{1}{|z|}$ provides the maximum value.

If we take $$a\ge \frac{1}{|z|}-|z|$$ we have $$|z|^2+a|z|-1\ge 0$$ which is not useful to find the maximum value of $|z|$.

• In the second question what I wanted to ask was, without calculating the actual values from the inequality how can we get to know which inequality provides the maximum value, because the book in its solution straightway discusses that as to find the max. value we considered $|z|-|1\z| \le a$ and did not give the argument as to why we did not choose the other inequality. So I was thinking maybe we could conclude the same from studying the nature of the curve of $x-(1/x)$ for $x \ge 0$ but I dont think I can conclude anything from that. Jul 16, 2016 at 14:26
• And, also in the answer you provided for the first part of the question you said that the book doesn't make the condition necessary but just to remove the trivial condition $0\lt |z|\lt 1$ it considered so, then why do we have to consider the case $|z| \lt 1$, for finding the least value of $a$. Jul 16, 2016 at 14:29
• @user350331: For the second, considering the graph should help. For the first, let me write my idea. For $0\lt |z|\le 1$, $a=|z+1/z|\gt 0\ge |z|-(1/|z|)$ holds for any $z$, so $|z|=1$ is the greatest value in this case. For $|z|\gt 1$, $(a+\sqrt{a^2+4})/2$ satisfies $|z|\gt 1$ for any $a\gt 0$, so $|z|=(a+\sqrt{a^2+4})/2$ is the greatest value. I think you have to "consider" the case $0\lt |z|\le 1$, i.e. what happens in that case, but you don't have to write the case if you know what happens, as the book omits. Jul 16, 2016 at 15:22
• So now I see as $|z+1/z|\ge |z|-|1/z|$ and also $|z+1/z|\ge |1/z|-|z|$, so to have a tighter bound on the inequality after considering $|z|\lessgtr 1$, we search for the maximum or minimum value whichever is required. Am I right. Also, if possible can you look into this one too Jul 16, 2016 at 15:51
• @user350331: Yes, you are right. You don't know how I've struggled to solve the linked question :) Jul 16, 2016 at 15:58