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 .

Therefore :

---------------------------------------------------------------------------------

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

---------------------------------------------------------------------------------

Now, develop (2').

Note that

Divide by 2 :

**Equating real and imaginary parts :** << here is the mistake... because b and c are complex numbers, it doesn't work. Pooh

From the second equation, we get

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 :

Substituting in (1'), one can see that this yields .

So b can't be a real number.

---------------------------------------------------------------------------------

**b is a complex number**

So :

[tex]b'=i \sqrt{3}[/tex]

:

If [tex]b'=i \sqrt{3}[/tex],

and

If [tex]b''=-i \sqrt{3}[/tex],

and

The couples of solutions are

We'll substitute in the equation (1') :

~~~~~~~~~~~~~~~~~~~~

First recall or note down a few useful things :

~~~~~~~~~~~~~~~~~~~~

Let's take the b',c' solutions.

We want this to be equal to -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..