Complex number inequality question

[bonbon22]

TL;DR

[z]<2 is a circle but im still not too sure why.

Z can be any point on the argand diagram so if z molous is less than 2 , is that somehow giving us the distance from origin? But how i assumed mod sign only makes things positive therefore its not sqrt( (x+yi)^2 ) = distance ??

 

[PeroK]

Summary: [z]<2 is a circle but I am still not too sure why.

Z can be any point on the argand diagram so if z molous is less than 2 , is that somehow giving us the distance from origin? But how i assumed mod sign only makes things positive therefore its not sqrt( (x+yi)^2 ) = distance ??

The modulus of a complex number $z$ is defined as $|z| = \sqrt{zz^*} = \sqrt{x^2 +y^2}$

 

[Mark44]

molous

You're missing a 'd'. The word is modulus.
For a complex number z, |z| is the distance from the origin to the point (x, y) in the complex plane.

 

[WWGD]

It's not a circle. Notice a circle is described by |z|=c , c a constant. You have an < instead.

 

[Mark44]

To elaborate on @WWGD's correction, the inequality $|z| < c$, with c a nonnegative real constant, is a disk. The boundary of a disk is a circle.