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

Question

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.

Answer 1

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 $u$.