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I'm trying to sketch $D=\{z \in \mathbb{C}: 2 \leq \vert z \vert < \vert z-2 \vert <4\}$.

I know that, geometrically, this is all $z$ whose distance from the origin is $\geq 2$ and is $<$ whose distance from $2$, which is less than $4$.

The question is: how do I sketch this information in the complex plane? Here's my attempt: the red represents the unwanted region and the green the desired region. Is this correct?

enter image description here Thanks

beep-boop
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  • The solution is not correct. You will get the region as the intersection of 2 annuli, one centered at 0, the other at 2 and the line $\text{Re}~z = 1$ – L__ May 09 '14 at 17:05

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Take the inequalities one at a time. The region $2 \le |z| < 4$ is an annulus, i.e. region between two circles, in this case including the inner circle of radius 2 centered at $0$ and excluding the outer circle of radius 4 (also centered at $0$).

The region $2 < |z-2| < 4$ is another annulus, this time not including either bounding circle, but centered at $2$. Finally the requirement $|z|<|z-2|$ describes the region to the left of the vertical line $x=1$ where $z=x+yi.$

Putting these together I got something that looks like two sections of "moons" which join at one point.

coffeemath
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  • Thanks! I've edited my original picture. Is it ok now? – beep-boop May 09 '14 at 21:03
  • You now need one more thing, namely to include only the points strictly to the left of the vertical line $x=1$. You also havd the Re and Im axes labels switched. The Re should be the horizontal and the Im should be vertical axes. – coffeemath May 10 '14 at 09:04