solutions of the system

**perash** · July 20th 2008, 11:01 PM

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 .

**Moo** · July 21st 2008, 01:58 AM

Hello,

Originally Posted by perash

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

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

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

~~~~~~~~~~~~~~~~~~~~
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..

**flyingsquirrel** · July 21st 2008, 03:58 AM

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 :

$ \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 $U,\,V$ and $W$ you'll get the values of $a,\,b$ and $c$ by solving

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

In case you wonder, I'm not sure this method will give you the result more quickly than Moo's one. Good luck !

**NonCommAlg** · July 21st 2008, 06:25 AM

Originally Posted by perash

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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