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.

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.
