Sketching a set of $2 < |z| \leq |z + 2| < 4$ on the complex plane.

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?

• Hint. $|z|$ is the distance of $z$ from the origin. $|z+2| = |z-(-2)|$ is the distance of $z$ from $-2$. Nov 7 '20 at 2:15
• Yeah, I know. I think I have just written that. Or you mean that I can use that fact to solve my problem (I mean this part: $|z| \leq |z + 2|$)? Nov 7 '20 at 2:18
• What points are closer to the origin than they are to $-2$? The closed half-plane$\Re(z)\geq-1$ Nov 7 '20 at 2:41
• @theman: Yes, exactly what saulspatz said. Nov 7 '20 at 8:05

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*}