Maximum and minimum absolute value of a complex number

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.

• 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). – 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?

• What a brilliant way to tackle this problem! I should have learnt a bit more Geometry particularly Circles. Any suggestion on that? – 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 $$z=2i$$.

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

• I like how you put it. I've been known to "lose myself in some computations" more times than I'd like – 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 $$\sqrt{\frac {9}{16}*4} =\frac 32$$.

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[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.