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 :
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Substituting in (1) and (2), we get the following system :
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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
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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.
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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') :
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First recall or note down a few useful things :
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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..