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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?

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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?

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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.

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

Just pretend it's

if i and 1 are

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To add to tiny-tim's hint: |b- a|, geometrically, is the distance between point a and b.

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Sorry, I've been staring at maths all weekend it's starting to all look the same.Are you sure [tex]z=2+2i[/tex] doesn't satisfy the inequality?

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]

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

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[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.

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Nice one, Mentallic!

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

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Darn this internet world. I'm detecting a hint of sarcasm, but can't be truly sureNice one, Mentallic!

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

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

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[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)

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Darn this internet world. I'm detecting a hint of sarcasm, but can't be truly sure

The 2nd line really threw me off.

my solution was a bit hand-wavy …

but yours is entirely analytic …

which in an analysis problem is actually quite appropriate!

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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.^{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)

Why just stop there? I'm also feeling depression, lack of self-esteem and gastric bowel troubles :tongue2:I'mdetecting paranoia …

my solution was a bit hand-wavy …

but yours is entirely analytic …

which in an analysis problem is actually quite appropriate!

No but honestly, I just felt that my method was so simple-minded that you were simply mocking it

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

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

the CAMpaign for REal Geometry

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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).

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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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