Maximum and minimum absolute value of a complex number

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?

$$f(z)=\left|z-\frac{1}{z}\right|=\frac{\left|z^2-1\right|}{2}$$ since we know $|z|=2$

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 $z=2i$.