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$|z| < |z-2i|$

Help. I know that each of the absolute values represents a circle with a radius that is unknown. However, I do not understand how to interpret the inequality and shade the corresponding region on the diagram.

enter image description here

Shooting Stars
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    The circles are, perhaps, not the most helpful picture here, since you don't really know the radius. You could try alternate methods, like substituting $z = x + iy$ (which should shake out rather nicely when you expand $|z|^2 = |z - 2i|^2$), or you could try thinking about the boundary of this region. Can you think of any points in $\Bbb{C}$ where $|z| = |z - 2i|$? The intersection of those circles give you two examples. What other points have this property? Which point on the imaginary axis has this property? If you can figure out the boundary, you just need to figure out which side to shade. – Theo Bendit Oct 06 '22 at 10:25

3 Answers3

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$|z|$ does not represent a circle centered at the origin, and $|z-2i|$ does not represent a circle centered at $2i$. Instead, $|z|$ represents the distance from $z$ to $0$, and $|z-2i|$ represents the distance from $z$ to $2i$.

Your inequality asks for which points $z$ in the plane the distance to $0$ is smaller than the distance to $2i$. In other words, which points in the plane are closer to $0$ than they are to $2i$?

Arthur
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  • For instance $|z| = 3$ does represent a circle centered at the origin with radius $3$, because it represents the points where the distance to the origin is $3$. But $|z|$ itself does not in any way represent a circle. – Arthur Oct 06 '22 at 10:31
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    +1. Your A appeared as I typed mine. I'm a very slow typist. – DanielWainfleet Oct 06 '22 at 10:44
  • The points that are closer to zero are the points below the midway point between $2i$ and $0$ so, $y=i$? – Shooting Stars Oct 06 '22 at 11:00
  • @ShootingStars I would say the points below the midway line, but otherwise, yes, exactly. – Arthur Oct 06 '22 at 11:01
  • Oh, I see, is it because of the definition of the line, that the set of points that are equidistant to two points, and this line is marked by $y=i$, correct? So the answer would be $y<i$? – Shooting Stars Oct 06 '22 at 11:04
  • @ShootingStars Essentially, yes, that's right. Notationally, you may want to be more specific about what $y$ is here. Assuming you're using the standard complex-number-decomposition $z=x+iy$, we actually write that solution as $y<1$, not $y<i$. Inequalities make little sense when one of the sides is non-real complex. – Arthur Oct 06 '22 at 11:13
  • I see, thanks @Arthur – Shooting Stars Oct 06 '22 at 12:10
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As the the result of the command of Mathematica 13.1

Reduce[Abs[z] < Abs[z - 2*I], z, Complexes]

Im[z]<1

shows, this is a half-plane of the complex plane.

user64494
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$|z|=|z-0|$ is the distance from $z$ to $0.$ And $|z+2i|=|z-(-2i)|$ is the distance from $z$ to $-2i.$ The right bisector of the line-segment that connects $0$ to $-2i$ is the line $\{u\in\Bbb C: Im(u)=-1\}.$ And $z$ is above this line, i.e. $Im(z)>-1,$ iff $z$ is closer to $0$ than to $-2i.$

DanielWainfleet
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