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

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

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 $$u$$.
• 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} – José Carlos Santos Oct 13 '19 at 16:19