Consider the complex numbers by real and imaginary parts.
A=a+ci
B=b+di
A2 = (a2 - c2) +(2ac)i
B2 = (b2 - d2) +(2bd)i
A3 = (a3 - 3ac2) + (3a2c - c3)i
A3 = (b3 - 3bd2) + (3b2d - d3)i
The system to solve is then
(a2 - c2) + (b2 - d2) = 7
(a3 - 3ac2) + (b3 - 3bd2) =10
(3a2c - c3) + (3b2d - d3) = 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:
2a2 - 2c2 = 7
2a3 - 6ac2 = 10
Solving for a (multiply the top by 3a and subtract the bottom)
gives the cubic 4a3 - 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.
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Posted by Jer
on 2015-12-14 11:07:40 |
Solution
