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## Homework Statement

Sketch all complex numbers

*z*which satisfy the given condition:

[itex]|z-i|\geq|z-1|[/itex]

## Homework Equations

[itex]z=a+bi[/itex]

[itex]|z|=\sqrt{a^{2}-b^{2}}[/itex]

## 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:

[itex]|(a+bi)-i|=|(a+bi)-1|[/itex]

[itex]\sqrt{(a)^{2}+(bi-i)^{2}}=\sqrt{(a-1)^{2}-b^{2}}[/itex]

[itex]a=-b[/itex]

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 [itex]\sqrt{-3}\geq\sqrt{3}[/itex] and [itex]\sqrt{1}\geq\sqrt{-1}[/itex]

What do I do from here?