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Math Help - complex equation

  1. #1
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    complex equation

    Im: (z+i)/(1-z) = 1
    |z|=1


    What i did so far:

    im: (a+bi+i)(1-a+bi)/(1-a-bi)(1-a+bi)=1
    im: (a-b-(a^2)-(b^2)-ai+bi+i)/((a^2)+(b^2)-2a+1)=1

    -a+b+1=(a^2)+(b^2)-2a+1
    (a^2)+(b^2)-a-b=0
    From here on forward i'm not sure how to proceed because i don't know how to implement the |z|=1 in the equation.What i did is this but i'm fairly certain it's not correct:
    (a-(1/2))^2+((b-(1/2))^2 - 1/2 = 0
    (a-(1/2))^2+((b-(1/2))^2 = (sqr.root 2/2)^2
    My assumption for the solution is: all points on radius of (sqr.root 2/2)^2 with center at (1/2,1/2)

    Would appreciate your help. Thank you
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  2. #2
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    Re: complex equation

    Quote Originally Posted by jakobjakob View Post
    Im: (z+i)/(1-z) = 1
    |z|=1
    What i did so far:
    im: (a+bi+i)(1-a+bi)/(1-a-bi)(1-a+bi)=1
    im: (a-b-(a^2)-(b^2)-ai+bi+i)/((a^2)+(b^2)-2a+1)=1
    The numerator is (a-a^2-b^2-b)+i(b-a+1)

    The denominator is (1-a)^2+b^2
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  3. #3
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    Re: complex equation

    here is my solution, which avoids "a's and b's".

    Given:

    \text{Im}\left(\frac{z+i}{1-z}\right) = 1, |z| = 1.

    the second equation tells us zz^* = 1.

    thus: z^* = \frac{1}{z}.

    now:

    \text{Im}\left(\frac{z+i}{1-z}\right) = \frac{1}{2i}\left[\frac{z+i}{1-z} - \left(\frac{z+i}{1-z}\right)^*\right]

     = \frac{1}{2i}\left[\frac{z+i}{1-z} - \frac{z^*-i}{1-z^*}\right]

     = \frac{1}{2i}\left[\frac{z+i}{1-z} - \frac{\frac{1}{z}-i}{1-\frac{1}{z}}\right] (since z* = 1/z)

    the fraction we are subtracting is a bit ugly, so let's simplify it, first:

    \frac{\frac{1}{z}-i}{1-\frac{1}{z}} = \frac{\frac{1-iz}{z}}{\frac{z-1}{z}} = \frac{1-iz}{z-1} = \frac{iz - 1}{1-z}

    oh look, we have a common denominator, now. so:

    \frac{1}{2i}\left[\frac{z+i}{1-z} - \frac{\frac{1}{z}-i}{1-\frac{1}{z}}\right] = \frac{1}{2i}\left[\frac{z+i-iz+1}{1-z}\right]

    now 1/i = -i. so we can replace 1/(2i) with -i/2. so we have:

    \frac{1}{2i}\left[\frac{z+i-1+iz}{1-z}\right] = \frac{(-i)(z+i-iz+1)}{2-2z} = \frac{z(-1-i) + (1-i)}{2-2z}

    remember, this whole mess is the imaginary part of (z+i)/(1-z), and is equal to 1. thus:

    z(-1-i) + (1-i) = 2 - 2z. this equation is linear in z, so we most certainly can solve it (over C). put all the z terms on one side, and the non-z terms on the other:

    z(-1-i+2) = 2-1+i. this easily simplifies to:

    z = \frac{1+i}{1-i}.

    but....we're not done, yet. let's simplify the fraction on the right:

    \frac{1}{1-i} = \frac{1+i}{2}, so:

    \frac{1+i}{1-i} = \frac{(1+i)^2}{2}.

    but:

    (1+i)^2 = 1 + 2i + i^2 = 2i.

    therefore:

    z = \frac{2i}{2} = i.

    verify that this indeed is a solution.

    (with all due respect for Plato, his method leads to one "false" solution, which must be eliminated: z = 1. this is due to "squaring introducing extraneous solutions").
    Thanks from jakobjakob and topsquark
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  4. #4
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    Re: complex equation

    I think i figured it out, but i would appreciate some feedback regarding correctness.

    (b-a+1)/((1-a)^2 + b^2) = 1 for all (a,b) =/= (1,0) (denominator can't be 0)

    b-a+1=(1-a)^2 + b^2
    a^2 + b^2 - a - b = 0

    |z|=1 = sqr.root (a^2 + b^2 )
    1=sqr.root (a^2 + b^2 ) /^2
    1=a^2 + b^2

    a^2 + b^2 - a - b = 0
    1 - a - b = 0
    b = 1 - a
    b = 1 - sqr.root(1 - b^2)
    sqr.root(1 - b^2)= 1 - b /^2
    1 - b^2 = 1 -2b + b^2
    2b^2 - 2b = 0
    b^2 - b = 0
    b(b - 1) = 0, b can't be 0 so the only solution is
    b = 1
    1 - a - b = 0
    1 - a - 1 = 0
    a = 0

    Solution:
    z=i
    Could this be correct?
    Edit: Ignore this post. I haven't seen Deveno's explanation until after i posted. Thank you Deveno.
    Last edited by jakobjakob; November 21st 2012 at 04:02 PM.
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  5. #5
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    Re: complex equation

    your solution is fine: you correctly identified the "extraneous solution" by realizing that b = 0 implies z = 1, which would make (z+i)/(1-z) undefined (and 1 certainly ISN'T undefined).

    the point is, we only have ONE possible z that works. i prefer to show such things by working with a single variable (z alone) rather than two (a and b), to avoid such difficulties (although i will admit, the algebra can get hairy this way).
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