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Thread: solutions of the system

  1. #1
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    solutions of the system

    Find all complex solutions of the system


    (a + ic)^3 + (ia + b)^3 + (−b + ic)^3 = −6 ,

    (a + ic)^2 + (ia + b)^2 + (−b + ic)^2 = 6 ,

    (1 + i)a + 2ic = 0 .
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  2. #2
    Moo
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    Hello,

    Quote Originally Posted by perash View Post
    Find all complex solutions of the system


    (a + ic)^3 + (ia + b)^3 + (−b + ic)^3 = −6 , (1)

    (a + ic)^2 + (ia + b)^2 + (−b + ic)^2 = 6 , (2)

    (1 + i)a + 2ic = 0 . (3)
    Let's study (3).

    Note that $\displaystyle 2i=(1+i)^2$.
    $\displaystyle (3) \Leftrightarrow (1+i)a+(1+i)^2c=0 \implies \boxed{a=-(1+i)c}$

    Therefore :
    $\displaystyle \begin{cases} a+ic=-c \\ ai+b=(1-i)c+b \end{cases}$

    ---------------------------------------------------------------------------------
    Substituting in (1) and (2), we get the following system :

    $\displaystyle \begin{cases} -c^3&+((1-i)c+b)^3+(-b+ic)^3=-6 \quad {\color{red}(1')} \\ c^2&+((1-i)c+b)^2+(-b+ic)^2=6 \quad {\color{red}(2')}\end{cases}$

    ---------------------------------------------------------------------------------
    Now, develop (2').

    Note that $\displaystyle (1-i)^2=-2i$
    $\displaystyle {\color{red}c^2}+(-2ic^2+b^2+2bc(1-i))+(b^2{\color{red}-c^2}-2ibc)=6$

    $\displaystyle -2ic^2+{\color{red}b^2}+2bc{\color{blue}-2ibc}+{\color{red}b^2}{\color{blue}-2ibc}=6$

    $\displaystyle 2b^2+2bc-2ic^2-4ibc=6$

    Divide by 2 :
    $\displaystyle (b^2+bc)-i(c^2+2bc)=3$


    Equating real and imaginary parts : << here is the mistake... because b and c are complex numbers, it doesn't work. Pooh
    $\displaystyle \begin{cases} b^2+bc=3 \\ c(c+2b)=0 \end{cases}$

    From the second equation, we get $\displaystyle c_1=0 \text{ and } c_2=-2b_2$

    $\displaystyle \begin{cases} b_1^2+b_1c_1=3 \\ b_2^2+b_2c_2=3 \end{cases}$

    $\displaystyle \begin{cases} b_1^2=3 \quad {\color{red}(3)} \\ -b_2^2=3 \quad {\color{red}(4)} \end{cases}$

    2 possibilities :
    - b is a real number
    - b is a complex number
    ---------------------------------------------------------------------------------
    b is a real number

    Therefore, the equation (4) is impossible and the possible solutions are :

    $\displaystyle \begin{cases} (c=0,b=\sqrt{3}) \implies a=0 \\ (c=0,b=-\sqrt{3}) \implies a=0 \end{cases}$

    $\displaystyle \boxed{\begin{cases} (a=0,b=\sqrt{3},c=0) \\ (a=0,b=-\sqrt{3},c=0) \end{cases}}$

    Substituting in (1'), one can see that this yields $\displaystyle 0=-6$.

    So b can't be a real number.

    ---------------------------------------------------------------------------------
    b is a complex number

    So $\displaystyle b_2=\pm i \sqrt{3}$ :
    [tex]b'=i \sqrt{3}[/Math]
    $\displaystyle b''=-i\sqrt{3}$

    $\displaystyle c_2=-2b_2$ :
    If [tex]b'=i \sqrt{3}[/Math], $\displaystyle c'=-2i \sqrt{3}$ and $\displaystyle a'=-(1+i)c'=2(-1+i) \sqrt{3}$

    If [tex]b''=-i \sqrt{3}[/Math], $\displaystyle c''=2i \sqrt{3}$ and $\displaystyle a''=-(1+i)c''=2(1-i) \sqrt{3}$


    The couples of solutions are

    $\displaystyle \boxed{\left\{\begin{array}{ccccc} a'=2(-1+i) \sqrt{3} & \quad & b'=i \sqrt{3} & \quad & c'=-2i \sqrt{3} \\
    & & & & \\
    a''=2(1-i) \sqrt{3} & \quad & b''=-i \sqrt{3} & \quad & c''=2i \sqrt{3}
    \end{array} \right.}$

    We'll substitute in the equation (1') :
    $\displaystyle -c^3+((1-i)c+b)^3+(-b+ic)^3=-6 \quad {\color{red}(1')}$

    ~~~~~~~~~~~~~~~~~~~~
    First recall or note down a few useful things :
    $\displaystyle i^3=-i$

    $\displaystyle (a+b)^3=a^3+b^3+3ab^2+3a^2b$

    $\displaystyle (1-i)^2=-2i$

    $\displaystyle (1-i)^3=-2i(1-i)=-2-2i=-2(1+i)$

    $\displaystyle (i \sqrt{3})^3=-3i \sqrt{3}$ (example)
    ~~~~~~~~~~~~~~~~~~~~
    Let's take the b',c' solutions.

    We want this to be equal to -6.
    $\displaystyle -(-2i \sqrt{3})^3+\underbrace{\left((1-i)(-2i \sqrt{3})+i \sqrt{3}\right)^3}_{(i \sqrt{3})^3 (-2(1-i)+1)^3=(i \sqrt{3})^3(-1+2i)^3}+\underbrace{(-i \sqrt{3}+i(-2i \sqrt{3}))^3}_{(\sqrt{3})^3(2-i)^3}$


    $\displaystyle 8(-3i \sqrt{3})+(-3i \sqrt{3})(-1+2i)^3+3 \sqrt{3}(2-i)^3$

    $\displaystyle -24i \sqrt{3}-3i \sqrt{3}(11-2i)+3 \sqrt{3}(2-11i)$

    $\displaystyle -24i \sqrt{3}-33i \sqrt{3}-6 \sqrt{3}+6 \sqrt{3}-33i \sqrt{3} \quad \boxed{\neq -6}$

    So the solution a',b',c' doesn't work.

    The same goes for a'',b'',c''.


    So there's no solution...
    I don't know if I'm wrong or if there is really no solution, but all this work for.. nothing, that's sad(istic)



    Edit : Ok, calculator gives a=1+i, b=2-i, c=-1 and a=1+i, b=-1-i, c=-1 and a=-2-2i, b=-1+2i, c=2
    I'll have a look later..
    Last edited by Moo; Jul 21st 2008 at 02:15 AM.
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  3. #3
    Super Member flyingsquirrel's Avatar
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    Hi

    I deleted my previous post because I found a mistake in my work but fixing it didn't change the method I used to solve the problem so here it is :

    Letting $\displaystyle U=a+ic$, $\displaystyle V=b+ia$ and $\displaystyle W=-b+ic$ the system becomes :

    $\displaystyle
    \begin{cases}
    U^3+V^3+W^3=-6\\
    U^2+V^2+W^2=6\\
    U+V+W=0\\
    \end{cases}$

    Once you've solve this for $\displaystyle U,\,V$ and $\displaystyle W$ you'll get the values of $\displaystyle a,\,b$ and $\displaystyle c$ by solving

    $\displaystyle \begin{cases}
    a+ic=U\\
    b+ia=V\\
    -b+ic=W
    \end{cases}$

    In case you wonder, I'm not sure this method will give you the result more quickly than Moo's one. Good luck !
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  4. #4
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    Quote Originally Posted by perash View Post
    Find all complex solutions of the system

    (a + ic)^3 + (ia + b)^3 + (−b + ic)^3 = −6 ,

    (a + ic)^2 + (ia + b)^2 + (−b + ic)^2 = 6 ,

    (1 + i)a + 2ic = 0 .
    flyingsquirrel's idea is good: put $\displaystyle u=a+ci, \ v=b+ai, \ w=-b+ci.$ then we'll have: $\displaystyle u+v+w=0, \ \ \ \ \ \ (1)$

    $\displaystyle u^2+v^2+w^2=6,$ and $\displaystyle u^3+v^3+w^3=-6.$ thus: $\displaystyle uv+uw+vw=\frac{(u+v+w)^2-u^2-v^2-w^2}{2}=-3, \ \ \ \ \ \ (2)$

    and: $\displaystyle uvw=\frac{u^3+v^3+w^3-(u+v+w)(u^2+v^2+w^2-uv-uw-vw)}{3}=-2. \ \ \ \ \ \ \ \ \ (3)$

    so by (1), (2), and (3), $\displaystyle u, v, w$ are the roots of $\displaystyle z^3-3z+2=0.$ but $\displaystyle z^3-3z+2=(z-1)^2(z+2).$ thus the roots

    are: $\displaystyle z=1, \ 1, \ -2.$ so there are 3 possibilities: $\displaystyle u=v=1, \ w=-2,$ or $\displaystyle u=w=1, \ v=-2,$ or $\displaystyle v=w=1, \ u=-2.$

    now since: $\displaystyle a=\frac{i+1}{2}(u-v-w),\ \ b=u-w-a, \ \ c=i(a-u),$ we'll have the following solutions:

    $\displaystyle u=v=1, \ w = -2,$ gives us: $\displaystyle a=i+1, \ b=2-i, \ c = -1.$

    $\displaystyle u=w=1, \ v = -2,$ gives us: $\displaystyle a=i+1, \ b=-i-1, \ c = -1.$

    $\displaystyle v=w=1, \ u=-2,$ gives us: $\displaystyle a=-2i-2, \ b=2i-1, \ c=2. \ \ \ \square$
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