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Complex equation with z

  1. Aug 16, 2011 #1
    Complex equation with "z"

    1. The problem statement, all variables and given/known data

    [tex]z^2-\bar z |z|= 1[/tex]

    2. Relevant equations

    Complex numbers

    3. The attempt at a solution


    [tex](a+ib)^2-(a-ib)\sqrt{a^2+b^2} = 1[/tex]

    [tex]a^2-b^2+2iab-a\sqrt{a^2+b^2}+ib\sqrt{a^2+b^2} = 1[/tex]

    I separate the real part from the imaginary part.

    [tex]
    \left\{\begin{matrix}
    2iab+ib\sqrt{a^2+b^2}=0

    \\
    a^2-b^2-a\sqrt{a^2+b^2}=1
    \end{matrix}\right.
    [/tex]

    One solution of the first is [itex]b=0[/itex]

    [tex]
    \left\{\begin{matrix}
    2a+\sqrt{a^2+b^2}=0, \ 3a^2=b^2

    \\
    a^2-b^2-a\sqrt{a^2+b^2}=1
    \end{matrix}\right.
    [/tex]

    So I basicalliy have two solution from the first eq.
    1) [itex]b=0[/itex]
    2)[itex] 3a^2=b^2[/itex]

    If you plug these results in the seconds I don't get anything...
    Either I get [itex]a^2-a^2=1[/itex] or [itex]-a^2=1[/itex] which has no solution since [itex]a \in \mathbb{R}[/itex]

    Could anyone confirm this equation has no solution. It's supposed to have one since it was part o an exam.
    Thanks.
     
  2. jcsd
  3. Aug 16, 2011 #2

    Ray Vickson

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    Science Advisor
    Homework Helper

    Re: Complex equation with "z"

    We can set up the equations for the real and imaginary parts as a polynomial system, then compute a Groebner basis. This will reveal *all* the solutions. Here it is done in Maple 14. First, let z = x + I*y (I = sqrt(-1) in Maple), to get x^2-y^2 + x*r -1 = 0 and 2xy-y*r=0, where y = sqrt(x^2+y^2). We can look at a Groebner basis for the polynomial system
    sys={2xy - r*y, x^2-y^2 + r*x - 1, r^2-x^2-y^2}. Maple:
    with(Groebner):
    B:=Basis(sys,tdeg(x,y,r));

    B := [y, x - r, 2 r^2 - 1]

    Therefore, a system equivalent to {sys1=0,sys2=0,sys3=0} is {B1=0,B2=0,B3=0}, or {y=0, x=r, r^2 = 1/2}.

    RGV
     
  4. Aug 16, 2011 #3
    Re: Complex equation with "z"

    Hi Ray, thanks for your time and help.

    I think I notice something not correct in your setup.
    You set
    x^2-y^2 + x*r -1 = 0
    as the real part
    and
    2xy-y*r=0
    as imaginary part,
    as I did.

    Notice that the original eq is [itex] z^2-\bar z |z|-1 = 0 [/itex]
    so I suppose your correct setup should be:
    x^2-y^2 - x*r -1 = 0
    2xy + y*r=0
    one sign is flipped over in each equation.


    I tried to understand as well
    B := [y, x - r, 2 r^2 - 1]

    Therefore, a system equivalent to {sys1=0,sys2=0,sys3=0} is {B1=0,B2=0,B3=0}, or {y=0, x=r, r^2 = 1/2}.

    RGV

    I supposed that last one is the solution. With the sign flipped over it may have solutions I suppose.
    I don't have Maple nor any evoluted math software, so if you don't mind supporting me a little bit more, please feed you software with correct eq.s and see what comes out, thanks !!! :smile:
     
    Last edited: Aug 16, 2011
  5. Aug 16, 2011 #4

    I like Serena

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    Homework Helper

    Re: Complex equation with "z"

    Hi Quinzio! :smile:

    In your OP you seem to be missing that [itex]\sqrt{a^2} = |a|[/itex].
    If you take that into account, you'll start finding solutions.
     
  6. Aug 16, 2011 #5
    Re: Complex equation with "z"

    Hello there ILS, thanks for the suggestion.
    [itex]a=\frac{-1}{\sqrt2}[/itex]

    is a solution
     
  7. Aug 16, 2011 #6

    I like Serena

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    Homework Helper

    Re: Complex equation with "z"

    Hmm.

    Your equation is:
    [tex]z^2 - \bar z |z| = 1[/tex]

    Your first solution is:
    [tex]z = a + i b = \frac {-1} {\sqrt 2}[/tex]

    Substituting I get:
    [tex](\frac {-1} {\sqrt 2})^2 - {\frac {-1} {\sqrt 2}} \cdot | \frac {-1} {\sqrt 2} |
    ~=~ \frac 1 2 - \frac {-1} {\sqrt 2} \cdot \frac {1} {\sqrt 2}
    ~=~ \frac 1 2 + \frac 1 2
    ~=~ 1[/tex]

    :wink:

    Edit: Hey, you corrected your post! :smile:
     
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