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Two different demonstrations on the complex plane

  1. Sep 3, 2011 #1
    Hi there, I have to prove this two sentences . I think I've solved the first, but I'm quiet stuck with the second.

    The first says:

    1) Demonstrate that the equation of a line or a circumference in the complex plane can be written this way: [tex]\alpha z . \bar{z}+\beta z+\bar{\beta} \bar{z}+\gamma=0[/tex], with [tex]\alpha,\gamma\in{R},\beta\in{C}[/tex]

    So I called z and beta:


    Then developing the products I get:

    And making alpha equal zero I get the equation for a line, right? (for u and v fixed).

    Then completing the square and reordering:

    [tex]\left(x+\displaystyle\frac{u}{\alpha}\right )^2+\left(y-\displaystyle\frac{v}{\alpha}\right )^2=\displaystyle\frac{u^2}{\alpha^2}+\frac{v^2}{\alpha^2}-\frac{\gamma}{\alpha}[/tex]

    This is the equation for the circle, is this right?

    In the other hand I got:

    2) Prove that the geometrical place for the points [tex]z\in{C}[/tex] that verifies [tex] \left\|{\displaystyle\frac{z-1}{z+1}}\right\|=k[/tex] is a circumference ([tex]k\in{R},1\neq{k}>0[/tex]).

    I couldn't make much for this. I called z=x+iy again:

    [tex] \left\|{\frac{z-1}{z+1}}\right\|=k\Rightarrow{\left\|{\frac{(x+iy-1)(x-iy+1)}{(x+iy+1)(x-iy+1)}}\right\|=k}\Rightarrow{\left\|{\frac{x^2+y^2+2iy-1}{(x+1)^2+y^2}}\right\|=k}[/tex]

    I don't know what to do from there.


    PS: sorry if it bothers you that I made two questions in the same topic, but I thought it was quiet trivial and it didn't deserved two different topics, but I can split it if necessary.
    Last edited: Sep 3, 2011
  2. jcsd
  3. Sep 3, 2011 #2
    The thing in the denominator is real by construction (You multiplied it by the complex conjugate). What are the real and imaginary parts of the numerator?
  4. Sep 3, 2011 #3

    BTW, is the first exercise right?
    Last edited: Sep 3, 2011
  5. Sep 3, 2011 #4
    The first exercise is correct.

    For the second one, a modulus of a complex number:
    |w| = \sqrt{ \left( \mathrm{Re} w \right)^{2} + \left( \mathrm{Im} w \right)^{2}} = k
    Square the equation, get rid off the denominators and simplify the expression. Write what you get.
  6. Sep 3, 2011 #5
    Thanks :)
  7. Sep 3, 2011 #6
    Actually, it does not seem to correspond to a circumference. The condition

    \left|\frac{z - 1}{z + 1}\right| = k \Leftrightarrow |z - 1| = k |z + 1|

    means that the distance from the point [itex](1, 0)[/itex] is [itex]k[/itex] times bigger than the distance from the point [itex](-1, 0)[/itex]. I am not sure if this is a circle in general.
  8. Sep 3, 2011 #7
    Actually, scratch what I was saying. Square the above:
    |z - 1|^{2} = k^{2} |z + 1|^{2}

    But, [itex]| w |^{2} = \bar{w} w[/itex], so:
    (z - 1)(\bar{z} - 1) = k^{2} (z + 1) (\bar{z} + 1)

    Simplify this and use part 1 of the problem.
  9. Sep 3, 2011 #8
    Thank you very much :)
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