If z^3+az^2+bz+c=0 is a polynomial over R with a complex root of the form z=α+αi, where α can be any real number show that: c^2=(b^2-2ac)(a^2-2b).
I cant seem to figure this out, can someone please help? Thanks
Or can someone give me a hint from where to start...
Things Ive tried: -Move c to the other side and square both sides, then make that equal to (b^2-2ac)(a^-2b), then hoped that everything cancels out and I am left with 0=0. :/
Given that P(z) is a polynomial over R defined by P(z) = z
3 + a z
2 + b z + c we have a, b, and c, are real [P is real if z is real]. Also, P(z
0)=0 where z
0=\(\displaystyle \alpha\,\,(1+i)\).
Since a, b, and c are real, if z is a complex root of P then z* is a root of P where * indicates conjugate, i.e. if P(z) = 0 then
P*(z)=(z*)
3 + a (z*)
2 + b (z*) + c = 0* = 0.
We also know that both the real and imaginary parts of P(z
0) and P(z
0*) are zero. So, what would you find out if you add and subtract P(z
0) and P(z
0*).
As an example, lets do the simpler case of P(z) = z
2 + a z + b, P(z
0)=0 where z
0=\(\displaystyle \alpha\,\,(1+i)\) and where a, b and \(\displaystyle \alpha\) are real. Prove 2 b + a
2 = 0.
Since \(\displaystyle z_0^2\, =\, \alpha^2\,\,(1+i)^2\, =\, 2\, \alpha^2\,\,i \),
P(z
0) = \(\displaystyle 2\, \alpha^2\, i\, +\, a\, \alpha\, (1+i) \, +\, b\, =\, 0\)
Similarly
P(z
0*) = \(\displaystyle -2\, \alpha^2\, i\, +\, a\, \alpha\, (1-i) \, +\, b\, =\, 0\)
Since the real and imaginary part of P(z
0) is zero we have
\(\displaystyle 2\, \alpha^2\, +\, a\, \alpha\, \, =\, 0\)
\(\displaystyle a\, \alpha\, +\, b\, =\, 0\)
Now, if \(\displaystyle \alpha\) were zero, the b would be zero and a root of P would be zero which is real. But P has two complex roots (z
0 and z
0*), so 0, which is real, can't be a root of P since P only has two roots. So \(\displaystyle \alpha\) is not zero and
\(\displaystyle a = - 2\, \alpha\)
\(\displaystyle b = - a \alpha = -\frac{a^2}{2}\)
or 2 b + a
2 = 0
Note that I didn't use P(z
0*) = 0 nor did I add or subtract P(z
0) and P(z
0*). You may (or may not) need that for your problem.