<s>[INST] 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.
 [/INST] $ 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 $\frac{\sqrt{9}}{2} = \frac{3}{2}$
</s>