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Math Help - Sketching regions in complex plane

  1. #1
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    Sketching regions in complex plane

    Draw a diagram showing the region {z \epsilon \mathbb{C} : Re(\frac{1}{z} - 2) \leq 1} ?

    Attempt:

    Re(\frac{1}{z} - 2) \leq 1

    Re(\frac{1 - 2z}{z}) \leq 1

    Re(\frac{1 - 2(x + iy)}{x + iy} ) \leq 1

    Re(\frac{1 - 2x - i2y}{x + iy} \frac{x - iy}{x - iy})

    Re(\frac{x - iy - 2x^2 - 2y^2}{x^2 + y^2}) \leq 1

    \frac{x - 2x^2 - 2y^2}{x^2 + y^2} \leq 1

    x - 2x^2 - 2y^2 \leq x^2 + y^2

    -3x^2 + x -3y^2 \leq 0

    3x^2 - x + \frac{1}{4} + 3y^2 \geq 4 (completing the square)

    I am not sure what else I can do
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  2. #2
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    Quote Originally Posted by SyNtHeSiS View Post
    Draw a diagram showing the region {z \epsilon \mathbb{C} : Re(\frac{1}{z} - 2) \leq 1} ?

    Attempt:

    Re(\frac{1}{z} - 2) \leq 1

    Re(\frac{1 - 2z}{z}) \leq 1

    Re(\frac{1 - 2(x + iy)}{x + iy} ) \leq 1

    Re(\frac{1 - 2x - i2y}{x + iy} \frac{x - iy}{x - iy})

    Re(\frac{x - iy - 2x^2 - 2y^2}{x^2 + y^2}) \leq 1

    \frac{x - 2x^2 - 2y^2}{x^2 + y^2} \leq 1

    x - 2x^2 - 2y^2 \leq x^2 + y^2

    -3x^2 + x -3y^2 \leq 0 Correct up to here.

    3x^2 - x + \frac{1}{4} + 3y^2 \geq 4 (completing the square)

    I am not sure what else I can do
    Complete the square a bit more carefully! First divide by 3, and get x^2 - \frac13x + y^2\geqslant0. Then you should get (x-\frac16)^2 + y^2 \geqslant\frac1{36}, which you ought to recognise as the region outside some circle.
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  3. #3
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    Looks good until the last line. I get (x-1/6)^2+y^2\ge (1/6)^2 after completing the square. This is the area outside the circle with the center (1/6, 0) and radius 1/6.
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