# Complex absolute value inequality

1. Oct 27, 2015

### Steve Turchin

$|z-1| + |z-5| < 4$

2. Relevant equations
$z = a + bi \\ |x+y| \leq |x| + |y|$ Triangle inequality
3. The attempt at a solution
$|z-1| + |z-5| < 4 \\ \\ x = z-1 \ \ , \ \ y = z-5 \\ \\ |z-1+z-5| \leq |z-1| + |z-5| \\ |2z-6| \leq |z-1| + |z-5| \lt 4 \ \ \Leftrightarrow \ \ |2z-6| \lt 4 \\ 2z-6 \lt 4 \ \ , \ \ -(2z-6) \lt 4 \\ z \lt 5 \ \ , \ \ \ \ \ \ \ \ \ \ \ \ -2z + 6 \lt 4 \ \ \ \Leftrightarrow \ \ \ -z \lt -2 \ \ \ \Leftrightarrow \ \ \ z \gt 1 \\ z \lt 5 \ \cap \ z \gt 1 \ \ \ \Leftrightarrow \ \ \ 1 \lt z \lt 5 \\ 1 \lt a+bi \lt 5 \ \ \Rightarrow \ \ \ 1 \lt a+bi \lt 5 \\ 1 \lt a \lt 5 \ \ \ \ \ , \ \ \ \ \ \ \ b = 0 \ \ \ \ for \ \ a,b \in \mathfrak R$
I think this is basically the interval $(1,5)$ on the Real axis.
I got this far, I doubt this is correct. Any tip on what a graphical representation of this would be?

2. Oct 27, 2015

### Staff: Mentor

I'm not sure how helpful the equations above are.
The inequality represents all of the points z in the complex plane for which the distance from 1 + 0i to z, plus the distance from z to 5 + 0i is less than 4. Can such a point be above or below the real axis? Why or why not?

3. Oct 27, 2015

### haruspex

A couple of problems with your analysis.
Your use of the triangle inequality allows you to make a deduction, but it also loses information. As a result, the region you end up with could be only part of the possible region. Secondly, you switch from complex to real, imposing further unjustified constraints. That's why you end up with just a line.
You'll get a much better idea if you start with the graphical view. |z-a| is the distance from z to point a, so your given condition is the sum of the distances from z to two given points. Does that remind you of any geometrical shape?