Can't solve this Compex numbers Problem? Pls help

Zhorsan

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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. :/
 

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View attachment 5140 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. :/
I think several here could help you, if you would bother to type out the question in a readable format.
 
I dont even know where to start with this question :/, someone please help...
 
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) = z3 + a z2 + b z + c we have a, b, and c, are real [P is real if z is real]. Also, P(z0)=0 where z0=\(\displaystyle \alpha\,\,(1+i)\).

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(z0) and P(z0*) are zero. So, what would you find out if you add and subtract P(z0) and P(z0*).

As an example, lets do the simpler case of P(z) = z2 + a z + b, P(z0)=0 where z0=\(\displaystyle \alpha\,\,(1+i)\) and where a, b and \(\displaystyle \alpha\) are real. Prove 2 b + a2 = 0.

Since \(\displaystyle z_0^2\, =\, \alpha^2\,\,(1+i)^2\, =\, 2\, \alpha^2\,\,i \),
P(z0) = \(\displaystyle 2\, \alpha^2\, i\, +\, a\, \alpha\, (1+i) \, +\, b\, =\, 0\)
Similarly
P(z0*) = \(\displaystyle -2\, \alpha^2\, i\, +\, a\, \alpha\, (1-i) \, +\, b\, =\, 0\)
Since the real and imaginary part of P(z0) 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 (z0 and z0*), 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 + a2 = 0

Note that I didn't use P(z0*) = 0 nor did I add or subtract P(z0) and P(z0*). You may (or may not) need that for your problem.
 
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