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Math Help - complex numbers: |z+1| + |z-1| = 3

  1. #1
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    [SOLVED] complex numbers: |z+1| + |z-1| = 3

    How can I solve an equation like this:
    <br />
|z+1| + |z-1| = 3, z \in \mathbb C<br />

    I've tried expanding z as
    <br />
z = x + iy; x,y \in \mathbb R<br />
    and computing the equation
    <br />
\sqrt{(x+1)^2 + y^2} + \sqrt{(x-1)^2 + y^2} = 3<br />
    but this way it quickly gets terribly complicated. Plus I have two variables in one expression.

    Any other ideas?
    Last edited by lampak; November 17th 2009 at 09:45 AM. Reason: solved
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by lampak View Post
    How can I solve an equation like this:
    <br />
|z+1| + |z-1| = 3, z \in \mathbb C<br />

    I've tried expanding z as
    <br />
z = x + iy; x,y \in \mathbb R<br />
    and computing the equation
    <br />
\sqrt{(x+1)^2 + y^2} + \sqrt{(x-1)^2 + y^2} = 3<br />
    but this way it quickly gets terribly complicated. Plus I have two variables in one expression.

    Any other ideas?
    Do you think there will be a finite number of solutions?
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  3. #3
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    Ah, no, there won't. I've forgotten that the task was "mark in the complex plane all numbers satisfying following condition".

    So maybe I don't need to calculate anything and just draw? But what? I have a feeling it will be an ellipse but I've never learnt about it.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by lampak View Post
    Ah, no, there won't. I've forgotten that the task was "mark in the complex plane all numbers satisfying following condition".

    So maybe I don't need to calculate anything and just draw? But what? I have a feeling it will be an ellipse but I've never learnt about it.
    It is indeed an ellipse.
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  5. #5
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    Quote Originally Posted by lampak View Post
    How can I solve an equation like this:
    <br />
|z+1| + |z-1| = 3, z \in \mathbb C<br />
    This is an ellipse in the complex plane: \text{Re}^2(z)+2\text{Im}^2(z)=1.
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  6. #6
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    But why, actually?
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  7. #7
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    Quote Originally Posted by lampak View Post
    Thanks
    But why, actually?
    If you want to avoid doing the algebra, then you can recognise it's an ellipse by noting that:

    1. geometrically, |z + 1| is the distance of z from -1.

    2. geometrically, |z - 1| is the distance of z from 1.

    3. geometrically, |z + 1| + |z - 1| = 3 means (distance of z from -1) + (distance of z from 1) = 3.

    4. The above is a well known locus definition of the ellipse (the foci are at z = 1 and z = -1). The cartesian of the ellipse is easily constructed from this: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 where 2a = 3 and 1^2 + b^2 = \left(\frac{3}{2}\right)^2 ....
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