3A. Find the smallest and largest value of the expression $\left|z-\frac{1}{z}\right|$, if $z$ is a complex number such that $|z|=2$.
Solution. First method. We have

$$
\left|z-\frac{1}{z}\right|=\left|\frac{z^{2}-1}{z}\right|=\frac{\left|z^{2}-1\right|}{|z|}=\frac{\left|z^{2}-1\right|}{2}
$$

Since

$$
\left|z^{2}-1\right| \geq\left|z^{2}\right|-1=4-1=3 \text { and }\left|z^{2}-1\right| \leq\left|z^{2}\right|+1=5
$$

we get $\frac{3}{2} \leq\left|z-\frac{1}{z}\right| \leq \frac{5}{2}$. If $z= \pm 2$, then $\left|z-\frac{1}{z}\right|=\frac{3}{2}$, so the minimum value of $\left|z-\frac{1}{z}\right|$ is $\frac{3}{2}$, and if $z= \pm 2 i$, then $\left|z-\frac{1}{z}\right|=\frac{5}{2}$, so the maximum value of $\left|z-\frac{1}{z}\right|$ is $\frac{5}{2}$.

Second method. Let $z=a+i b$. From $|z|=2$ it follows that $a^{2}+b^{2}=4$, so we have:

$$
\begin{aligned}
\left|z-\frac{1}{z}\right|^{2} & =\left|\frac{z^{2}-1}{z}\right|^{2}=\frac{\left|z^{2}-1\right|^{2}}{|z|^{2}}=\frac{\left|z^{2}-1\right|^{2}}{4}=\frac{1}{4}|z-1|^{2}|z+1|^{2} \\
& =\frac{1}{4}\left(a^{2}-2 a+1+b^{2}\right)\left(a^{2}+2 a+1+b^{2}\right) \\
& =\frac{1}{4}(5-2 a)(5+2 a)=\frac{1}{4}\left(25-4 a^{2}\right) \\
& =\frac{1}{4}\left(25-4\left(4-b^{2}\right)\right)=b^{2}+\frac{9}{4}
\end{aligned}
$$

From $a^{2}+b^{2}=4$ it follows that $0 \leq b^{2} \leq 4$, so $\frac{3}{2} \leq\left|z-\frac{1}{z}\right| \leq \frac{5}{2}$. The minimum value is for $z= \pm 2$ and is $\frac{3}{2}$, and the maximum value is for $z= \pm 2 i$ and is $\frac{5}{2}$.