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Math Help - Complex number equation

  1. #1
    Junior Member Evales's Avatar
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    Complex number equation

    Question:
    The complex number (z = x + yi) satisfies the the equation |z-3|+|z+3|=10

    What are the two points on the real axis which satisfy this equation?

    What are the two points of the imaginary axis which satisfy this equation?

    Hence sketch the locus of z on this diagram.

    My answer:
    For the two real points I got -5 and 5.
    Because I ended up with |5| = z
    Since z = x + yi I thought that y = 0


    My queries:
    But looking down apparently I was wrong because it then asks for the value of y.

    Anybody able to point me in the right direction?

    Btw, what is a locus?
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  2. #2
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    Quote Originally Posted by Evales View Post
    Question:
    The complex number (z = x + yi) satisfies the the equation |z-3|+|z+3|=10

    What are the two points on the real axis which satisfy this equation?

    What are the two points of the imaginary axis which satisfy this equation?

    Hence sketch the locus of z on this diagram.

    My answer:
    For the two real points I got -5 and 5.
    Because I ended up with |5| = z
    Since z = x + yi I thought that y = 0


    My queries:
    But looking down apparently I was wrong because it then asks for the value of y.

    Anybody able to point me in the right direction?

    Btw, what is a locus?
    A locus is a set of points that satisfies a given relationship. You want the set of points in the complex plane that satisfy |z-3|+|z+3|=10.

    Geometrically, this is the set of points z such that the distance of z from z = 3 plus the distance of z from z = -3 is equal to 10. Geometrically this defines an ellipse. The points z = 3 and z = -3 are the focal points. The length of the major axis is 10. By symmetry the centre is at (0, 0).

    Therefore, if z = x + iy, then the Cartesian equation has the form \frac{x^2}{5^2} + \frac{y^2}{b^2}= 1.

    To get b, you can apply Pythagoras Theorem:

    Since the focii are at (3, 0) and (-3, 0), 3^2 + b^2 = (10/2)^2 \Rightarrow b = 4.

    Therefore the Cartesian equation is \frac{x^2}{5^2} + \frac{y^2}{4^2}= 1.


    Of course, you can do it the hard way by substituting z = x + iy into |z-3|+|z+3|=10 and going through the algebra. If I have time later I'll post the highlights.
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  3. #3
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    Quote Originally Posted by Evales View Post
    Question:
    The complex number (z = x + yi) satisfies the the equation |z-3|+|z+3|=10

    What are the two points on the real axis which satisfy this equation?

    What are the two points of the imaginary axis which satisfy this equation?

    Hence sketch the locus of z on this diagram.

    My answer:
    For the two real points I got -5 and 5.
    Because I ended up with |5| = z
    Since z = x + yi I thought that y = 0


    My queries:
    But looking down apparently I was wrong because it then asks for the value of y.

    Anybody able to point me in the right direction?

    Btw, what is a locus?
    By the way, without using the Cartesian equation I derived, the values of z on the imaginary (y) axis that satisfy the given relationship are found by substituting x = 0:

    | iy - 3 | + | iy + 3| = 10 \Rightarrow 2 \sqrt{y^2 + 9} = 10 \Rightarrow y^2 + 9 = 5^2 \Rightarrow y = \pm 4, as expected.
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