For all complex numbers with $|z|=2$, inequality $2\le |z-4|\le 6$ holds

Question

Prove for all $|z| = 2$, $$2 \leq |z - 4| \leq 6$$

I tried $|(x - 4) + iy| = |x^2 + 16 - 8x + y^2| = |20 - 8x|$

I also tried using triangle inequality $2 = |z| = |z - 4 + 4| \leq |z - 4| + 4$

I can't bound the above.

Answer 1

$2=|4|-|z|\leq |4-z|\leq |z|+|4|=6$, using the inequality $|a|-|b|\leq |a-b|$ wich is equivalent to $|a|=|a-b+b|\leq |a-b|+|a|$.

Answer 2

Think of it geometrically: given that $z$ is a point on the circle centered at $0+0i$ with radius $2$, the question asks you to show that $z$ lies inside an annulus centered at $4+0i$ with inner radius $2$ and outer radius $6$:

Answer 3

WLOG we can write $z=2(\cos\theta+i\sin\theta)$ where $\theta$ is real

For real $a,$

$$|z-a|^2=(2\cos\theta-a)^2+(2\sin\theta)^2=4+a^2-4a\cos\theta$$

If $a\ge0,$

$\displaystyle-1\le\cos\theta\le1$ will imply $\displaystyle(a-2)^2\le4+a^2-4a\cos\theta\le(a+2)^2$

$$\iff a-2\le|z-a|\le a+2$$