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Math Help - complex number equation

  1. #1
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    complex number equation

    I need to solve the following equation and i am encountering some problem

    Re: (z+i)/(1-z)=0

    This is what i did so far:
    Re:
    (z+i)/(1-z) +1 = 1
    (1+i)/(z-1)=-1
    .
    and then i multiply left side with (z(conjugate)-1)/(z(conjugate)-1).
    After all the calculations i get the following equation and i don't know which a or b value is correct:
    a(a+1)+b(b+1)=0

    z=complex number, a= Re value of complex number, b=Im value of complex number

    I hope i've expressed myself clear and thank you for your help.
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: complex number equation

    For the given equation, this implies z+i=0.
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  3. #3
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    Re: complex number equation

    Quote Originally Posted by jakobjakob View Post
    I need to solve the following equation and i am encountering some problem
    Re: (z+i)/(1-z)=0
    I think the actual problem is \text{Re}\left(\frac{z+i}{1-z}\right)=0

    Note that \text{Re}\left(\frac{z}{w}\right)=\frac{\text{Re}(  z)\text{Re}(w)+\text{Im}(z)\text{Im}(w)}{|w|^2}

    If you want that to be 0, then the numerator must be zero.

    Also z=-i~\&~0 are two solutions.
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  4. #4
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    Re: complex number equation

    How could z = -i possibly be a solution to this equation? The real parts of a complex number are always real...

    \displaystyle \begin{align*} \textrm{Re}\left( \frac{z + i}{1 - z} \right) &= 0 \\ \textrm{Re}\left[ \frac{x + y\,i + i}{1 - \left( x + y\,i \right)} \right] &= 0 \\ \textrm{Re}\left[ \frac{x + \left( y + 1 \right) i}{1 - x - y\,i} \right] &= 0 \\ \textrm{Re}\left\{ \frac{\left[ x + \left( y+ 1\right) i \right]\left( 1 - x + y\,i \right)}{\left( 1 - x - y\, i \right) \left( 1 - x + y\, i\right) } \right\} &= 0 \\ \textrm{Re}\left[ \frac{x(1 - x)  + xy\,i + (1 - x)\left( y + 1 \right) i + y\left( y + 1 \right) i^2}{(1 - x)^2 + y^2 }  \right] &= 0 \\ \textrm{Re}\left\{ \frac{x(1 - x) - y(y + 1) + \left[ xy + (1 - x)(y + 1) \right] i }{(1 - x)^2 + y^2} \right\} &= 0 \\ \textrm{Re} \left\{ \frac{x(1 - x) - y(y + 1)}{(1 - x)^2 + y^2} + \left[\frac{xy + (1 - x)(y + 1)}{(1 - x)^2 + y^2}\right] i \right\} &= 0 \end{align*}

    \displaystyle \begin{align*} \frac{x(1 - x) - y(y + 1)}{(1 - x)^2 + y^2} &= 0 \\ x(1 - x) - y(y + 1) &= 0 \textrm{ for all }(x, y) \neq (1, 0) \\ x - x^2 - y^2 - y &= 0 \\ x^2 - x + y^2 + y &= 0 \\ x^2 - x + \left( -\frac{1}{2} \right)^2 + y^2 + y + \left( \frac{1}{2} \right)^2 &= \left( -\frac{1}{2} \right)^2 + \left( \frac{1}{2} \right)^2 \\ \left( x - \frac{1}{2} \right)^2 + \left( y + \frac{1}{2} \right)^2 &= \left( \frac{\sqrt{2}}{2} \right)^2 \end{align*}


    So the solution to your problem is all points that lie on the circle of radius \displaystyle \begin{align*} \frac{\sqrt{2}}{2} \end{align*} units, centred at \displaystyle \begin{align*} \left( \frac{1}{2}, -\frac{1}{2} \right) \end{align*}.
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