So I need to sketch $ 2 < |z| \leq |z + 2| < 4 $ on the complex plane. At first, it seamed pretty easy since I know that:
- $2 < |z|$ is a circle with a center in point $0,0$ and radius = 2
- |z + 2| < 4 is a circle with a center in point $-2,0$ and radius = 4
BUT here comes that part $ |z| \leq |z + 2| $ and I actually have no idea how to connect those two.
Edit – my proposed solution:
$ |z| \leq |z + 2| \implies z^2 \leq z^2 + 4z + 4 \implies 0 = \leq 4z + 4 \implies -1 \leq z $
Is that correct? Is it the missing condition?
Best Answer
HINT
You cannot square both sides as if $z$ were real. Indeed, if we set $z = x + yi$, one has that \begin{align*} |z| \leq |z+2| & \Longleftrightarrow |z|^{2} \leq |z+2|^{2}\\\\ & \Longleftrightarrow z\overline{z} \leq (z+2)(\overline{z} + 2)\\\\ & \Longleftrightarrow z\overline{z} \leq z\overline{z} + 2(z+\overline{z}) + 4\\\\ & \Longleftrightarrow 0 \leq 4x + 4\\\\ & \Longleftrightarrow x\geq -1 \end{align*}