if z is complex root of x^5+3x^3-2x^2+10, prove conjugate is

[DarkSun]

will be glad to get a hint regarding how it can be done...

p=x^5+3x^3-2x^2+10 polynomical with real coefficients.
prove that if z is complex root of p, then z conjugate is also a root.

 

[fasteddie65]

I would substitute z = a + bi into the equation and see if you get 0.
Then substitute z* = a - bi also, and see if you also get 0.
If you don't get 0 for both, you should get the same value at least.

Or you could solve the equation using possible rational roots. As I look at the graph on my calculator, I see only one real root, and it is NOT rational. So using synthetic division to reduce the degree is not an option.

I guess I would go back to my original idea.

 

[pka]

\(\displaystyle p=x^5+3x^3-2x^2+10\)
Polynomial with real coefficients.
Prove that if z is complex root of p, then z conjugate is also a root.

Suppose that in addition to z, each of a, b, c, & d is also a root.
Then by the factor theorem: \(\displaystyle p=(x-z) (x-a) (x-b) (x-c) (x-d)\).
That means that the constant term \(\displaystyle 10= z \cdot a \cdot b \cdot c \cdot d\).
If each of a, b, c, & d is real then the constant is complex. BUT it is 10, a real number.
So one of those must be complex, say a, so that \(\displaystyle z \cdot a \text{ is real}.\)
But the only complex numbers times z that gives a real number is a multiple of \(\displaystyle \overline z\).

 

[mmm4444bot]

I would substitute z = a + bi into the equation and see if you get 0. <<< There is no way to get zero when working with symbols.

If we substitute x = a + bi into the equation, then we get:

p = a^5 + 5i a^4 b - 10 a^3 b^2 - 10i a^2 b^3 + 5 a b^4 + i b^5 + 3 a^3 + 9i a^2 b - 9 a b^2 - 3i b^3 - 2a^2 - 4i a b + 2 b^2 + 10

Is the right-hand side of this equation equal to zero? It is, but only after substituting one of the four unknown pair of Real values for a and b that make it so.

Then substitute z* = a - bi also, and see if you also get 0.
If you don't get 0 for both, you should get the same value at least.

Again, you won't get any values. You won't even get the same expression.

I would approach this exercise by writing a proof that covers all polynomials, instead.

Any decent math text on the topic will have such a proof; this proof also exists on the Interet at tens of thousands of locations.

If my suggestion doesn't toot your scooter, then maybe you can get away with using decimal approximations of an actual Complex root.

Let x = -0.4513603517 - 2.024554649i

Now show that you get the same value for p with both x and x*.

 

[pka]

Re:

I would approach this exercise by writing a proof that covers all polynomials, instead.
Any decent math text on the topic will have such a proof; this proof also exists on the Interet at tens of thousands of locations.

Do you understand that this was posted in the Beginning Algebra Forum?
The real proof is ridiculously simple: \(\displaystyle 0 = \overline 0 = \overline {P(z)} = P\left( {\overline z } \right)\).

 

[DarkSun]

Thanks, got it.
The 'general' proof is really much simpler than a 'specific' proof here.

 

[mmm4444bot]

Do you understand that this was posted in the Beginning Algebra Forum?

Yes, I do.

Is that always supposed to mean the same thing?