Consider the complex numbers by real and imaginary parts.
A=a+ci
B=b+di
A^{2} = (a^{2}  c^{2}) +(2ac)i
B^{2} = (b^{2}  d^{2}) +(2bd)i
A^{3} = (a^{3}  3ac^{2}) + (3a^{2}c  c^{3})i
A^{3} = (b^{3}  3bd^{2}) + (3b^{2}d  d^{3})i
The system to solve is then
(a^{2}  c^{2}) + (b^{2}  d^{2}) = 7
(a^{3}  3ac^{2}) + (b^{3}  3bd^{2}) =10
(3a^{2}c  c^{3}) + (3b^{2}d  d^{3}) = 0
If A+B is real we also know c=d *see below
and with the second equation of the system we also get a=b
Substitute these into the first and third equations to get the simpler system:
2a^{2}  2c^{2} = 7
2a^{3}  6ac^{2} = 10
Solving for a (multiply the top by 3a and subtract the bottom)
gives the cubic 4a^{3}  21a + 10 = 0
which has roots 2.5, .5, 0
The complex number solutions to the original system are then
A = 2.5 + isqrt(11)/2, B = 2.5  isqrt(11)/2, A+B=5
A = .5  sqrt(13)/2, B = .5 + sqrt(13)/3, A+B=1
(note they are real)
A = 2 + isqrt(1/2), B = 2 = isqrt(1/2), A+B=4
*I didn't know what to make of "real value of A+B" so I took it as a hint. These solutions seem to work but the above does not rule out other solutions where A+B is complex but their real parts sum to more than 4.

Posted by Jer
on 20151214 11:07:40 