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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?

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  • $\begingroup$ Hint. $|z|$ is the distance of $z$ from the origin. $|z+2| = |z-(-2)|$ is the distance of $z$ from $-2$. $\endgroup$
    – Brian Tung
    Nov 7 '20 at 2:15
  • $\begingroup$ 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: $|z| \leq |z + 2|$)? $\endgroup$
    – theman
    Nov 7 '20 at 2:18
  • $\begingroup$ What points are closer to the origin than they are to $-2$? The closed half-plane$\Re(z)\geq-1$ $\endgroup$
    – saulspatz
    Nov 7 '20 at 2:41
  • $\begingroup$ @theman: Yes, exactly what saulspatz said. $\endgroup$
    – Brian Tung
    Nov 7 '20 at 8:05
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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*}

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  • $\begingroup$ Great, that is the hint that I needed. I got right answer and I wouldn't have known that I did an error on my way. Thank you very much Sir! $\endgroup$
    – theman
    Nov 7 '20 at 2:55
  • $\begingroup$ You are welcome! I am glad to help. $\endgroup$
    – user0102
    Nov 7 '20 at 2:56

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