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Math Help - Sketching |z - 1| + |z + i| <= 2 in complex plane

  1. #1
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    Sketching |z - 1| + |z + i| <= 2 in complex plane

    Sketch the following region in the complex plane: { z \epsilon\mathbb{C} : |z -1| + |z + i| \leq 2}

    Attempt:

    |z - 1| + |z + i| \leq 2

    \sqrt{(x - 1)^2 + y^2} + \sqrt{x^2 + (y + 1)^2} \leq 2

    \sqrt{(x - 1)^2 + y^2} \leq 2 - \sqrt{x^2 + (y + 1)^2}

    x^2 - 2x + 1 + y^2 \leq 4 - 4\sqrt{x^2 + y^2 + 2y + 1} + (x^2 + y^2 + 2y + 1)

    -2x - 4 - 2y \leq -4\sqrt{x^2 + y^2 + 2y + 1}

    (-2x - 4 - 2y)^2 \leq 16(x^2 + y^2 + 2y + 1)

    4x^2 + 16x + 16y + 8xy + 16 + 4y^2 \leq 16(x^2 + y^2 + 2y + 1)

    It looks like I have reached a dead end
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  2. #2
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    |z-1| is the distance of z to 1.
    |z+i| is the distance of z to -i.
    So you have the sum of two is less than or equal 2.
    Think ellipse.
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  3. #3
    MHF Contributor Also sprach Zarathustra's Avatar
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    Or: -12x^2+8xy+16 x-12y^2-16y <0

    Which is an ellipse!
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  4. #4
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    Quote Originally Posted by SyNtHeSiS View Post
    Sketch the following region in the complex plane: { z \epsilon\mathbb{C} : |z -1| + |z + i| \leq 2}

    Attempt:

    |z - 1| + |z + i| \leq 2

    \sqrt{(x - 1)^2 + y^2} + \sqrt{x^2 + (y + 1)^2} \leq 2

    \sqrt{(x - 1)^2 + y^2} \leq 2 - \sqrt{x^2 + (y + 1)^2}

    x^2 - 2x + 1 + y^2 \leq 4 - 4\sqrt{x^2 + y^2 + 2y + 1} + (x^2 + y^2 + 2y + 1)

    -2x - 4 - 2y \leq -4\sqrt{x^2 + y^2 + 2y + 1}

    (-2x - 4 - 2y)^2 \leq 16(x^2 + y^2 + 2y + 1)

    4x^2 + 16x + 16y + 8xy + 16 + 4y^2 \leq 16(x^2 + y^2 + 2y + 1)

    It looks like I have reached a dead end
    The boundary of that set is given by 4x^2 + 16x + 16y + 8xy + 16 + 4y^2 = 16(x^2 + y^2 + 2y + 1)
    which, as Plato and also Sprach Zarathustra have told you, is an ellipse. The set you want is the interior and boundary of that set (because of the "<").
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  5. #5
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    Thanks. How would you change

    -12x^2 + 16x - 12y^2 - 16y + 8xy \leq 0

    into the standard form on an ellipse:

    \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 ?
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  6. #6
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    The quick answer is: one does not.
    The axes are not vertical/horizontal.
    The foci are (1,0) and (0,-1)
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  7. #7
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    In that case, how would you know that it is an ellipse, with only knowing those 2 points (1,0) and (0, -1)?
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  8. #8
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    Quote Originally Posted by SyNtHeSiS View Post
    In that case, how would you know that it is an ellipse, with only knowing those 2 points (1,0) and (0, -1)?
    ELLIPSE
    The path of a point which moves so that the sum of its distances from two fixed points is a constant. The two fixed points are the foci of the ellipse.
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  9. #9
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    The internal region is where the sum is <2.

    Sorry about the typo on the sketch, should be |z+i|.
    Attached Thumbnails Attached Thumbnails Sketching |z - 1| + |z + i| &lt;= 2 in complex plane-ellipse.jpg  
    Last edited by Archie Meade; October 27th 2010 at 11:48 AM.
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  10. #10
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    Thanks. I am still a bit confused. I understand that the definition of an ellipse is that the sum of the distances from both foci to a point is constant, but how would you know this in this example, without testing 2 points on the ellipse, to see if the sum of their distances from both foci are equal?
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  11. #11
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    Quote Originally Posted by SyNtHeSiS View Post
    Thanks. I am still a bit confused. I understand that the definition of an ellipse is that the sum of the distances from both foci to a point is constant, but how would you know this in this example, without testing 2 points on the ellipse, to see if the sum of their distances from both foci are equal?
    The notation \left| {z - z_0 } \right| is the distance from z to z_0.

    Therefore \left| {z - z_0 } \right|+\left| {z - z_1 } \right|=C must be an ellipse.
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  12. #12
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    Quote Originally Posted by Plato View Post
    The notation \left| {z - z_0 } \right| is the distance from z to z_0.

    Therefore \left| {z - z_0 } \right|+\left| {z - z_1 } \right|=C must be an ellipse.
    Provided C > \left|z_1 - z_0\right|.

    @OP: If C = \left|z_1 - z_0\right| then the locus is a line segment joining z_0 and z_1, and if C < \left|z_1 - z_0\right| then there is no locus.
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