Let, $z \in \mathbb C$ and $|z|=2$. What's the maximum and minimum value of $$\left| z - \frac{1}{z} \right|? $$

I only have a vague idea to attack this problem. Here's my thinking :

Let $z=a+bi$

Exploiting the fact that, $a^2+b^2=4$

We get $z-\dfrac{1}{z}=a-\dfrac{a}{4}+i\left(b+\dfrac{b}{4}\right)$

So $$ \begin{split} \left|z-\frac{1}{z}\right| &=\sqrt{\left(a-\dfrac{a}{4}\right)^2+\left(b+\dfrac{b}{4}\right)^2}\\ &=\sqrt{4+\dfrac{1}{4}-\dfrac{a^2}{2}+\dfrac{b^2}{2}}\\ &=\sqrt{4+\dfrac{1}{4}+\dfrac{1}{2}(b^2-a^2)} \end{split} $$

The minimum value can be obtained if we can minimize $b^2-a^2$. Setting $b=0$ gives the minimum value $\sqrt{2+\dfrac{1}{4}}=\dfrac{3}{2}$

Now, comes the maximum value. We can write $$\sqrt{4+\dfrac{1}{4}+\dfrac{1}{2}(b^2-a^2)}$$ $$=\sqrt{4+\dfrac{1}{4}+\dfrac{1}{2}(4-2a^2)}$$ $$=\sqrt{4+\dfrac{1}{4}+2-a^2}$$ $$=\sqrt{6+\dfrac{1}{4}-a^2}$$

Setting $a=0$ gives the maximum value $\sqrt{6+\dfrac{1}{4}}=\dfrac{5}{2}$.

I don't know if it's okay to set $b=0$ since $z$ would become a real number then.

  • 1
    $\begingroup$ The real numbers are a subset of the complex numbers. That a possible values of $z$ has an imaginary part of $0$ (a.k.a. $0i$), does not make it non-complex. This is much like the way that all integers are also rationals (and reals, and complex). $\endgroup$ – John Bollinger Oct 10 '18 at 1:45

A (very) faster way:

We know that $z$ is in the circle centered at $0$ and radius $2$ and $1/z$ is in the circle of center $0$ and radius $1/2$. The maximum distance between a point of the former and a point of the latter is $$2+\frac12=\frac52$$ Now, we need to show that there exists some $z$ such that this distance is reached. Take $z=2i$.

Can you deal with the minimum now?

  • $\begingroup$ What a brilliant way to tackle this problem! I should have learnt a bit more Geometry particularly Circles. Any suggestion on that? $\endgroup$ – ARahman Oct 10 '18 at 14:39

$$f(z)=\left|z-\frac{1}{z}\right|=\frac{\left|z^2-1\right|}{2}$$ since we know $|z|=2$


You could try like this (with help of triangle inequality):

$$|z-{1\over z}|= |{z^2-1\over z}| = {|z^2-1|\over 2} \geq {|z^2|-1\over 2} ={3\over 2}$$

clearly this can be achieved at $z = 2$ and $$ {|z^2-1|\over 2} \leq {|z^2|+1\over 2} = {5\over 2}$$ which can be achieved at $z=2i$.

  • $\begingroup$ The maximum isn't achieved at $z=-2$, though, is it? Since $|z - 1/z| = |-2-1/(-2)| = |-2+1/2| = 3/2$. $\endgroup$ – David Z Oct 10 '18 at 7:06
  • $\begingroup$ Yeah, thanks, I corrected it. $\endgroup$ – Aqua Oct 10 '18 at 7:53

Your idea is good, but you lose yourself in some computations (the maximum is correct, though).

Consider the square of the modulus: $$ f(z)=\left|z-\frac{1}{z}\right|^2=\frac{|z^2-1|^2}{|z|^2} $$ Since $|z|=2$ by assumption, we can as well consider $$ g(z)=|z^2-1|^2=(z^2-1)(\bar{z}^2-1)=z^2\bar{z}^2-z^2-\bar{z}^2+1=5-z^2-\bar{z}^2 $$ If $z=a+bi$, then $z^2=a^2-b^2+2abi$ and $\bar{z}^2=a^2-b^2-2abi$; but you also know that $a^2+b^2=4$, so $b^2=4-a^2$. Then $$ g(a+bi)=5-2a^2+2b^2=5-2a^2+20-2a^2=25-4a^2 $$ The maximum is for $a=0$, the minimum for $a=\pm2$; thus the maximum value of $g$ is $25$ and the minimum value is $9$.

We just need to divide by $4$ and take the square root, so the maximum for $f$ is $5/2$ and the minimum is $3/2$.

  • $\begingroup$ I like how you put it. I've been known to "lose myself in some computations" more times than I'd like $\endgroup$ – Yuriy S Oct 10 '18 at 2:18

$ z - \frac{1}{z} = \frac{z^2-1}{z} $

So that, we get $\left| z - \frac{1}{z} \right| $ = $\frac{|z^2-1|}{2} $

If we put $z=x+iy$, then $z^2 = x^2 - y^2 + 2ixy $, and we get:

$ \frac{|x^2-y^2-1 + i(2xy)|}{2} = \frac{\sqrt{(x^2-y^2-1)^2 + (2xy)^2}}{2} $

Let's take a look at function: $ f(x,y) = (x^2-y^2-1)^2 + 4x^2y^2 $

We know, that $|z| = 2$, so $x^2+y^2 = 4 $, and by that $ y^2 = 4-x^2, x \in [-2,2] $

$(x^2 - 4 + x^2 - 1)^2 + 4x^2(4-x^2) = (2x^2-5)^2 + 16x^2 - 4x^4 = \\ = 4x^4 - 20x^2 + 25 + 16x^2 - 4x^4 = 25-4x^2 $

So, all we have to do is looking at $25-4x^2, for \ x^2 \in[0,4] $ and say what's the min and max.

Clearly if $x=0$ then we get $25$ and the max value is $\frac{\sqrt{25}}{2} = \frac{5}{2}$

if $x^2 = 4$ we get $9$ and the min value is $\frac{\sqrt{9}}{2} = \frac{3}{2}$


Hint: $z - \frac 1z = z - \frac {\overline z}{z\overline z} = z - \frac {\overline z}{|z|^2} = z - \frac {\overline z}4 = \frac 34 Re(z) -i\frac 54 Im(z)$[1]

Let $Re(z) = a$ and $Im(z) = b$ which can be any values so that $a^2 + b^2 = 4$.

So you are being asked to find the maximum and minimum possible values of $\sqrt {\frac {9}{16}a^2 + \frac {25}{16}b^2}$ given that $a^2 + b^2 = 4$.

Intuitively[2] I'd say that as $\frac {25}{16} > \frac {9}{16}$ and given that $a^2=4-b^2$ increases/decreases when $b^2$ decreases/increases, this will be max when $b^2$ is max and $a^2$ is least and min when $a^2$ is max and $b^2$ is min.

As $a^2, b^2 \ge 0$ then $a^2$ is max when $b^2 = 0$ and $a^2 = 4$ and $a^2$ is least when $a^2 = 0$ and $b^2 = 4$.

So max is $\sqrt {\frac {25}{16}*4} = \frac 52$ and min is $\sqrt{\frac {9}{16}*4} =\frac 32$.


[1]. There probably no need for this, but we can create an identity:

$z \pm \frac 1z = (1\pm \frac 1{|z|^2})Re(z) + i(1 \mp \frac 1{|z|^2})Im(z)$ for any $z\ne 0$.

[2] And formally I'd say $\sqrt {\frac {9}{16}a^2 + \frac {25}{16}b^2} =\sqrt {\frac {25}{16}b^2 + \frac {9}{16}(4-b^2)}=\sqrt{b^2 + \frac 94}$ which is max/min when $b$ is max/min.

Actually formally was easier than my intuition.


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