Math Is Fun Forum

thedarktiger

Member

Registered: 2014-01-10

Posts: 91

Polynomial eugh

Let x, y, and z be complex (i.e., real or nonreal) numbers such that x+y+z, xy+xz+yz, and xyz are all positive real numbers.

Is it necessarily true that x, y, and z are all real, and positive? If so, prove it. If not, give a counterexample.

Thanks. I can prove that they are positive if they are real numbers, but don't see the approach for if they are complex.

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Bob

Administrator

Registered: 2010-06-20

Posts: 10,621

Re: Polynomial eugh

hi thedarktiger,

Here's my initial thoughts on this:

So this is a polynomial with real coefficents => one root must be real, and the other two must be complex conjugates.

So you could try making up a pair of complex conjugates, add in a real root and check to see if this provides a counter example as suggested.

Bob

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thedarktiger

Member

Registered: 2014-01-10

Posts: 91

Re: Polynomial eugh

I found that 3i, -3i, and 2 work so theres one.
Why does one root have to be real when the coefficients are?

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anonimnystefy

Real Member

From: Harlan's World

Registered: 2011-05-23

Posts: 16,049

Re: Polynomial eugh

It has to do with the nature of the roots of real polynomials. When you have a polynomial with real coefficients, if it has a complex root, the the conjugate of that root is also a root. So, you have three roots, and all must be complex conjugates of some root. So, what can happen is:
1. We have three real roots, each being its own complex conjugate;
2. We have 2 complex root, which means the third one must be its own complex conjugate, and therefore real.

This logic works for any odd number of roots.

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Olinguito

Member

Registered: 2014-08-12

Posts: 649

Re: Polynomial eugh

I found that 3i, -3i, and 2 work so theres one.

Any numbers of the form[list=*]
[*]

[/*]
[/list]where a, b, c are real will work.

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