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Thread: Complex numbers

  1. #1
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    Complex numbers

    1. If $\displaystyle z=x+iy$ satisfies $\displaystyle \mid z-4 \mid + \mid z+4 \mid =10$ show that

      $\displaystyle (x/5)^2+(y/3)^2=1$


    2. Conversely, if x and y satisfy $\displaystyle (x/5)^2+(y/3)^2=1$, then show that $\displaystyle z=x+iy$ satisfies $\displaystyle \mid z-4 \mid + \mid z+4 \mid =10$.



    Not a clue how to do this, nothing seems to work.
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  2. #2
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    These may help you. You can square both sides.
    $\displaystyle \left| {z - 4} \right|^2 = \left( {z - 4} \right)\left( {\overline z - 4} \right) = z\overline z - 4z - 4\overline z + 16$

    $\displaystyle \left| {z + 4} \right|^2 = \left( {z + 4} \right)\left( {\overline z + 4} \right) = z\overline z + 4z + 4\overline z + 16$

    $\displaystyle \left| {z - 4} \right|^2 + \left| {z + 4} \right|^2 = 2z\overline z + 32$

    $\displaystyle \left( {z + 4} \right)\left( {z - 4} \right) = z^2 - 16$
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  3. #3
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    Hello, Thomas!

    Another approach . . .


    If $\displaystyle z\:=\:x+iy$ satisfies $\displaystyle |z-4| + |z + 4|\:=\:10$

    show that: .$\displaystyle \left(\frac{x}{5}\right)^2 + \left(\frac{y}{3}\right)^2\:=\:1$
    $\displaystyle |z-4|$ is the distance from $\displaystyle z$ to the point (4,0).

    $\displaystyle |z+4|$ is the distance from $\displaystyle z$ to the point (-4,0).

    If the sum of the distances to two fixed points is a constant,
    . . the locus of $\displaystyle z$ is an ellipse, whose equation can be derived
    . . with the Distance Formula (and a lot of algebra).

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  4. #4
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    Thanks. I can do (1) by taking $\displaystyle \left| z+4 \right|$ to be $\displaystyle \sqrt{(x+4)^2+y^2}$ and the same for the other. However I'm having trouble doing the opposite, starting with the ellipse eqn and finding that $\displaystyle z=x+iy$...
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