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

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

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

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

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

    2b^2+2bc-2ic^2-4ibc=6

    Divide by 2 :
    (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
    \begin{cases} b^2+bc=3 \\ c(c+2b)=0 \end{cases}

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

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

    \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 :

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

    \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 0=-6.

    So b can't be a real number.

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

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

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

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


    The couples of solutions are

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

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

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

    (a+b)^3=a^3+b^3+3ab^2+3a^2b

    (1-i)^2=-2i

    (1-i)^3=-2i(1-i)=-2-2i=-2(1+i)

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

    We want this to be equal to -6.
    -(-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}


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

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

    -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; July 21st 2008 at 03:15 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 U=a+ic, V=b+ia and W=-b+ic the system becomes :

    <br />
\begin{cases}<br />
U^3+V^3+W^3=-6\\<br />
U^2+V^2+W^2=6\\<br />
U+V+W=0\\<br />
\end{cases}

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

    \begin{cases}<br />
a+ic=U\\<br />
b+ia=V\\<br />
-b+ic=W<br />
\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 u=a+ci, \ v=b+ai, \ w=-b+ci. then we'll have: u+v+w=0, \ \ \ \ \ \ (1)

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

    and: 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), u, v, w are the roots of z^3-3z+2=0. but z^3-3z+2=(z-1)^2(z+2). thus the roots

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

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

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

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

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