Sketching inequalities involving complex numbers

[phosgene]

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?

 

[Infinitum]

|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

 

[phosgene]

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! :)

 

[Infinitum]

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}