# Sketching inequalities involving complex numbers

## Homework Statement

Sketch all complex numbers z which satisfy the given condition:

$|z-i|\geq|z-1|$

## Homework Equations

$z=a+bi$

$|z|=\sqrt{a^{2}-b^{2}}$

## The Attempt at a Solution

First I find the boundary between the regions where the inequality holds and does not hold by replacing the inequality sign with an equality sign. Then I substitute z=a+bi into the equation and solve for b:

$|(a+bi)-i|=|(a+bi)-1|$

$\sqrt{(a)^{2}+(bi-i)^{2}}=\sqrt{(a-1)^{2}-b^{2}}$

$a=-b$

So the boundary cuts through the origin and has a slope of -1.

To find which side of the boundary the equality holds, I can plug random values from different sides of the boundary, in this case z=-1-i and z=1+i.

Doing this, I get $\sqrt{-3}\geq\sqrt{3}$ and $\sqrt{1}\geq\sqrt{-1}$

What do I do from here?

$|z|=\sqrt{a^{2}-b^{2}}$

This is incorrect. $|z| = \sqrt{a^2+b^2}$

You can also solve this without using a and b, in a rather simple way. This how I would try, by drawing the graph. $|z-i|$ represents all circles(vectors) with center at (0,1) and $|z-1|$ represents circles with center (1,0). The perpendicular bisector of the line joining these two points gives the condition when $|z-i|=|z-1|$. So, which side of the curve would you sketch z, such that $|z-i|>|z-1|$?

Try drawing a graph, and it should become really simple to figure out

Ah, I thought i was included in the formula. I re-did the entire thing with the correct form, and I got it now. Thanks! :)

Ah, I thought i was included in the formula. I re-did the entire thing with the correct form, and I got it now. Thanks! :)

Graphically, modulus/absolute value represents the magnitude of distance between the point and origin. Keeping that in mind, its easy to see why it comes out to be $\sqrt{a^2+b^2}$