How Do You Graph the Complex Inequality |z-1|<|z|?

[torquerotates]

Homework Statement

Graph the inequality: |z-1|<|z| where z=x+iy {i is the imaginary number: (-1)^.5}

Homework Equations

for complex #'s z and w,
|w+z|<or=|w|+|z|
|z-w|>or=|z|-|w|

The Attempt at a Solution

|z|-|1|<or=|z-1|<|z| { if we consider 1 to be complex i.e 1=1+0i}
=>|z|-1<|z|
=>-1<0

I have no idea how to graph this last inequality. Isn't it just a true statement in general?

 

[gabbagabbahey]

Homework Equations

for complex #'s z and w,
|w+z|<or=|w|+|z|
|z-w|>or=|z|-|w|

Huh? What do these inequalities (which are always true) have to do with the inequality in the question?

Instead, use $z=x+iy$ to calculate both $|z|$ and $|z-1|$ in terms of x and y and then substitute your results into the given inequality.

 

[HallsofIvy]

Also, one can interpret |x- y|, geometrically, as the distance form x to y in the complex plane. That means that we can think of |z- 1| as the distance from z to 1+ 0i or from (x,y) to (1, 0) and |z| as the distance from z to 0+0i or from (x,y) to (0,0). Saying that |z-1|< |z| means the point (x,y) is closer to (1, 0) than to (0,0).
The line x= 0.5 (the complex numbers 0.5+ yi for any real number y) is the perpendicular bisector of the interval from (0,0) to (1,0). That line is the set of points z so that |z-1|= |z|. The points for which |z-1|< |z| is the set of points to the right of that line.