Graphing a complex inequality

  • #1

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?
 

Answers and Replies

  • #2
gabbagabbahey
Homework Helper
Gold Member
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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 [itex]z=x+iy[/itex] to calculate both [itex]|z|[/itex] and [itex]|z-1|[/itex] in terms of x and y and then substitute your results into the given inequality.
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
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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.
 

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