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I'm trying to figure out the following set:

$|z+i| = 2|z|$.

My approach was to use the triangle inequality as follows:

$2|z| = |z+i| \leq |z| + |i| = |z| + 1$. This should imply that

$|z| \leq 1$, but this is nowhwere near the solution.

lhoernle
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5 Answers5

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Hint. The given condition means that the distance from $z$ to $(-i)$ is twice the distance from $z$ to $0$. It is well known that the locus of points having a given ratio of distances to two given points is a circle. Such circles are called Apollonian circles. See e.g. this article at cut-the-knot.org.

Alex
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Observe \begin{align} |z+i|^2=4|z|^2. \end{align} Since we have that \begin{align} |x+iy+i|^2=x^2+(y+1)^2 = 4x^2+4y^2=4|z|^2 \end{align} which means \begin{align} 3x^2+3\left(y^2-\frac{2}{3}y+\frac{1}{9}\right)-\frac{1}{3}=3x^2+3y^2-2y=1 \ \ \ \Rightarrow \ \ \ 3x^2+3\left(y-\frac{1}{3} \right)^2 = \frac{4}{3}. \end{align}

Jacky Chong
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Let $z=x+iy$, $|z|=\sqrt{x^2+y^2}$ $$x^2+(y+1)^2 = 4x^2+4y^2$$ $$3x^2+3y^2-2y-1=0$$ $$x^2+(y-\frac{1}{3})^2=(\frac{2}{3})^2$$ $$|z - (0+\frac{1}{3}i )| =|z-\frac{1}{3}i |=\frac{2}{3}$$

It is a circle.

Jay Zha
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  • I understand your solution and the identical one above from Jacky Chong and appreciate it ... but it seems I should be able to visualize this in terms of z and the definition of circles in the complex plane. – lhoernle Apr 14 '17 at 04:34
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    @lhoernle I updated the answer in terms of complex number, is that what you are expecting to see? – Jay Zha Apr 14 '17 at 04:44
  • Yes, more along these lines ... I guess I don't see how to go directly from the definition of the set to $|z-\frac{1}{3}| = \frac{2}{3}$, without going through the equation of a circle in the Euclidean plane. Is there a way of doing so? – lhoernle Apr 14 '17 at 04:48
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    @lhoernle I think using the (x,y) representation is the most natural way for it though. And you could think all of this is happening on complex plain (though it is most equivalent to Euclidean plain here). Think x is for the real line, and y for the imaginary line. And all operation is in complex plain. And we just expand z into (x,y) representation, and put it back to z finally using the definition of |z| – Jay Zha Apr 14 '17 at 04:53
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$|z+i|^2 = (2|z|)^2$

$z = x + i y$

$x^2 + (y+1)^2 = 4x^2 + 4y^2$

$3x^2 + 3y^2 - 2y -1 =0$

$x^2 + y^2 -\frac{2}{3}y - \frac{1}{3} = 0$

It is form of a circle if we compare it with the general equation $x^2 + y^2 + 2gx + 2fy + c =0$ whose center is $(-g,-f) = (0,\frac{1}{3})$ and radius of $\sqrt{g^2 + f^2 -c} = \sqrt{4/9} = \frac{2}{3}$.

You see that as per your approach too here $|z| < 1$.

More on properties and general equations of circles - http://mathscentre.ac.uk/resources/workbooks/mathcentre/mc-TY-circles-2009-1.pdf

BAYMAX
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Here |z+i|=2|z|. Let us assume z=x+iy. thus |x+i(y+1)|=2|x+iy|.

(x^2+(y+1)^2)^(1/2)= 2(x^2+y^2)^(1/2).

x^2 + y^2 +2y +1= 4(x^2+y^2).

3(x^2+y^2)-2y-1=0. Which is an equation of circle in complex plain with centre(0, 1/3) and radius 2/3.

Suprabha
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  • What's essentially the same answer has been posted several times already. – dxiv Apr 14 '17 at 04:45
  • Other answers also have same meaning. – Suprabha Apr 14 '17 at 04:55
  • Yes, indeed, but they were all posted within minutes of each other. Yours came noticeably later and, honestly, adds nothing new to what had been already said. Sorry, but that's a downvote in my books. – dxiv Apr 14 '17 at 05:08