Sketching inequalities involving complex numbers

  • Thread starter phosgene
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  • #1
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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?
 

Answers and Replies

  • #2
881
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[itex]|z|=\sqrt{a^{2}-b^{2}}[/itex]

This is incorrect. [itex]|z| = \sqrt{a^2+b^2}[/itex]


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

Try drawing a graph, and it should become really simple to figure out :smile:
 
  • #3
144
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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! :)
 
  • #4
881
40
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 [itex]\sqrt{a^2+b^2}[/itex]
 

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