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Thread: Two different demonstrations on the complex plane

  1. #1
    Senior Member
    May 2010

    [SOLVED] Two different demonstrations on the complex plane

    EDIT: Its done

    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: \alpha z . \bar{z}+\beta z+\bar{\beta} \bar{z}+\gamma=0, with \alpha,\gamma\in{R},\beta\in{C}

    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:

    \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}

    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 z\in{C} that verifies  \left\|{\displaystyle\frac{z-1}{z+1}}\right\|=k is a circumference ( k\in{R},1\neq{k}>0).

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

     \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}

    I don't know what to do from there.


    (Sorry if this topic doesn't belong here, please move it, I didn't know where to post so I posted here, and sorry too 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 by Ulysses; Sep 3rd 2011 at 04:08 PM. Reason: Solved.
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  2. #2
    Junior Member
    Jan 2011

    Re: Two different demonstrations on the complex plane

    I was getting ready to type up a nice geometric solution to the second part that I found in Heath's translation of the Elements and I saw that you solved it. I think it's still interesting enough to add here, although I'll leave the details to anyone interested. It shows that the locus of a point such that its distances from two given points are in a given ratio (not equal to one) is a circle. In your case, you want all points z whose distances from 1 and -1 are in the ratio k

    The following proposition can be extended to the case where the external angle EAC is bisected and meets line BC produced.

    Now if B and C are the two given points (1 and -1 in this case), there are two points X and Y on line BC that satisfy what we're after, one on segment BC and one outside segment BC. Construct the circle with XY as diameter and the extended proposition above can be used to prove that you have precisely the points in the complex plane that satisfy the condition of your second problem.
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