Given $|z-2+i| \le 2$ find minimum and maximum of $|z|$

Question

Given $|z-2+i| \le 2$ find minimum and maximum of $|z|$

Using trianlge inequality, $|z-(2-i)| \ge |z|-|i-2| $ or $|z| \le \sqrt 5 + 2$

and $|z+(i-2)| \le |z| + |i-2|$ but then I cant say $|z | + \sqrt 5 \ge 2$. How can I get the answer $\sqrt5 - 2\le|z|\le\sqrt 5 + 2$ ?

Answer 1

$||z| - |2-i|| \le |z - 2 + i|\le 2 \Rightarrow -2 \le |z| - \sqrt 5 \le 2$

Answer 2

Geometrically you are looking for the points on the circle with radius $2$ $$|z-z_0|^2 \leq 4 \mbox{ with } z_0 = 2-i \Rightarrow |z_0|^2 = 5$$ with minimal and maximal distance to the origin of the complex plane.

$$|z|^2 = |z_0 + (z-z_0)|^2 = |z_0|^2 + |z-z_0|^2 + 2 Re(\overline{z_0}(z-z_0)) = $$ $$ = 5 + r^2 + 2\cdot 5 \cdot r \cos{(\angle{(z_0,z-z_0))}} \mbox{ where } 0 \leq r=|z-z_0| \leq 2$$

So, for any $0 \leq r \leq 2$: $$ (r-\sqrt{5})^2 \leq |z|^2 \leq (r+\sqrt{5})^2$$ Minimum and maximum is reached simultaneously on both sides for $r=2$ where $\angle{(z_0,z-z_0))} = \pi$ for the minimum and $\angle{(z_0,z-z_0))} = 0$ for the maximum.

Answer 3

This is easy to visualise, $z$ must be inside of circle (or on it) with radius $2$ and center at $w=2-i$ Now when will $|z|$ be max or min? Clearly when $0,z$ and $w$ are collinear.