Plotting complex inequalities

Trying to plot the set of complex numbers [tex]\{z \in C | |z - i| \leq |z-1|\} [/tex] on the complex plane.

I've tried by hand a few values for z, such as [tex]z = 1 - 2i[/tex], [tex]z = 2 + 2i[/tex], but the inequality isn't true.

How can I determine which values of z are true for this inequality?
 

Mentallic

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Are you sure [tex]z=2+2i[/tex] doesn't satisfy the inequality? :wink:

If you let [tex]z=x+iy[/tex] and remember that [tex]|x+iy|=\sqrt{x^2+y^2}[/tex] you should be able to get your answer in terms of x and y.
 

tiny-tim

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Trying to plot the set of complex numbers [tex]\{z \in C | |z - i| \leq |z-1|\} [/tex] on the complex plane.
Hi username12345! :smile:

Just pretend it's geometry

if i and 1 are points, what words would you use to describe, geometrically, |z - i| ≤ |z - 1|? :wink:
 

HallsofIvy

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To add to tiny-tim's hint: |b- a|, geometrically, is the distance between point a and b.
 
Are you sure [tex]z=2+2i[/tex] doesn't satisfy the inequality? :wink:
Sorry, I've been staring at maths all weekend it's starting to all look the same.

Yeah it will be true because the displacement is the same on both axis. [tex]\sqrt(2^2 + 1^2) = \sqrt(1^2 + 2^2)[/tex] :smile:

I will have to return to this problem later, my head hurts. Thanks for your help everyone.
 

Mentallic

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I don't like to oppose a mentor's suggestion, but I try to avoid geometry whenever possible, especially when algebra provides a perfectly feasible solution. If you don't feel the same way, of course, just ignore my ramblings. (geometry in this case would be simpler, but that is only if you know where you are going with it. Algebra is a simple, fool proof step-by-step process)

[tex]\{z \in C | |z - i| \leq |z-1|\} [/tex]

So if we let [tex]z=x+iy[/tex] and substitute into the original question: [tex]|x+iy-i| \leq |x+iy-1|[/tex]
and collect real and imaginary terms: [tex]|x+i(y-1)| \leq |(x-1)+iy|[/tex]

Remember that: [tex]|a+ib|=\sqrt{a^2+b^2}[/tex]

So now if we follow this rule, we have: [tex]\sqrt{x^2+(y-1)^2} \leq \sqrt{(x-1)^2+y^2}[/tex]

and now you can simplify this inequality in terms of x and y. This will give you the cartesian inequality, but it is equivalent and can be copy-pasted onto the argand diagram.
 

tiny-tim

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Nice one, Mentallic! :smile:

and you can use squares throughout, to avoid the √ at the end :wink:
 

Mentallic

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Nice one, Mentallic! :smile:

and you can use squares throughout, to avoid the √ at the end :wink:
Darn this internet world. I'm detecting a hint of sarcasm, but can't be truly sure o:)
The 2nd line really threw me off.

I know my recommended solution is silly and inferior compared to the simple geometry of this problem, but it's these silly moves that keep it interesting :smile:
 

Office_Shredder

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Mentallic, the second line just means noting |a|<|b| if and only if |a|2<|b|2 so your last line becomes

[tex]x^2+(y-1)^2} \leq \sqrt{(x-1)^2+y^2[/tex]

It's fairly common when comparing absolute values to square them first because the algebra is always nicer at the end (in this case, you can just square both sides after getting rid of the absolute value signs, but it's good to know as a general rule)
 

tiny-tim

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boo!

Darn this internet world. I'm detecting a hint of sarcasm, but can't be truly sure o:)
The 2nd line really threw me off.
I'm detecting paranoia :wink: …​

my solution was a bit hand-wavy …

but yours is entirely analytic …

which in an analysis problem is actually quite appropriate! :biggrin:
 

Mentallic

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Mentallic, the second line just means noting |a|<|b| if and only if |a|2<|b|2 so your last line becomes

[tex]x^2+(y-1)^2} \leq \sqrt{(x-1)^2+y^2[/tex]

It's fairly common when comparing absolute values to square them first because the algebra is always nicer at the end (in this case, you can just square both sides after getting rid of the absolute value signs, but it's good to know as a general rule)
Oh yeah. Since x and y are real, and the square root suggests they are positive, squaring both sides of the inequality is no problem.

I'm detecting paranoia :wink: …​

my solution was a bit hand-wavy …

but yours is entirely analytic …

which in an analysis problem is actually quite appropriate! :biggrin:
Why just stop there? I'm also feeling depression, lack of self-esteem and gastric bowel troubles :tongue2:
No but honestly, I just felt that my method was so simple-minded that you were simply mocking it :wink:

I'm glad you agree and please, sign a petition to ban the use of geometry :biggrin:
 

tiny-tim

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support CAMREG!

I'm glad you agree and please, sign a petition to ban the use of geometry :biggrin:
oooh no … i like to sneak geometry in wherever i can! :biggrin:

support CAMREG!
the CAMpaign for REal Geometry :approve:
 

matt grime

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Geometry to understand and the algebra to compute ought to be the moral.

It may also help to plot the equality first, then decide which of the two regions the plane is separated into corresponds to the given inequality (the other part of the plane will correspond to the reverse inequality).
 

HallsofIvy

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Here's my geometry solution: |z- i| is the distance from the point z= x+ iy and the point i. |z-1| is the distance from the point z= x+ iy to the point 1. The set of points for which |z- i|= |z- 1| is the perpendicular bisector of the line segment between i and 1. The set of points for which [itex]|z-i|\le |z- 1|[/itex] is the set of points above and to the left of that bisector. That should be sufficient to plot the set which was what was required.

If you want more than that, go to Cartesian geometry: the segment from (0, 1) to (1, 0) has midpoint (1/2, 1/2) and slope -1. The perpendicular must have slope 1 and pass through (1/2, 1/2): y= x. The set of points such that [itex]|z-i|\le|z-1|[/itex] is [itex]\{x+ iy | x\le y\}[/itex].

I don't believe all that algebra was simper or more accurate than that!
 

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