<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] 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 $3/2$.
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