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.
 A: 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?
A: $$f(z)=\left|z-\frac{1}{z}\right|=\frac{\left|z^2-1\right|}{2}$$
since we know $|z|=2$
A: 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 $z=2i$.
A: 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 $3/2$.
A: $ 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 $\frac{\sqrt{9}}{2} = \frac{3}{2}$
A: 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.
