I have to draw a region that satisfies inequalities $|z|<2$ and $|z-u|<|z|$ where $u=-\sqrt{3}+i$ on the Argand diagram

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I drew these on cartesian plane by expanding

this method is very long if I expand and simplify

But how to draw second inequality

Is there any way ?

Marking scheme says that draw a line from orgin to $u$ then take perpendicular bisector of it.


The condition $\lvert z-u\rvert<\lvert z\rvert$ simply means that $z$ is closer to $u$ than to $0$. So, consider the perpendicular bisector of the line segment joining $u$ to $0$. It divides $\mathbb C$ into two half-planes. Now, take the half-plane that contains $u$.

  • $\begingroup$ BTW, can we simplify this inequality in any way ?? $\endgroup$ – AKA Death Oct 13 '19 at 16:15
  • $\begingroup$ I would write $z$ as $x+yi$. Then\begin{align}\lvert z-u\rvert<\lvert z\rvert&\iff\left(x+\sqrt3\right)^2+(y-1)^2<x^2+y^2\\&\iff2\sqrt3x+3-2y+1<0\\&\iff\sqrt3x-y<-2.\end{align} $\endgroup$ – José Carlos Santos Oct 13 '19 at 16:19

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