0
$\begingroup$

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

enter image description here

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.

$\endgroup$
2
$\begingroup$

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 $u$.

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.