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Question about roots in polynomial p(X)

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eecs4ever

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Question about roots

Suppose a polynomial p(x) has real coeficients.

and if z1 is a root of p(x) . z1 may be complex.
prove the conjugate of z1 is also a root.

if z1 is real, then z1 conjugate = z1, and the proof is complete.
How about if z1 is complex?
 

jayc

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Re: Question about roots

If the root z1 is complex and the coefficients of the polynomial are all real, you must somehow get rid of the complex part of the root when it is multiplied by other roots.

for example, if f(x) has complex roots z1, z2, z3, ...:

f(x) = (x-z1)*(x-z2)*....

Just multiplying out the first 2 terms, we get

f(x) = (x^2 - z2 * x - z1 * x + z1 * z2)*....

For the polynomial coefficients to be real, it must be true that

(z2 + z1) is real and z1*z2 is real. This can only be achieved if z2 and z1 are complex conjugates.

It is also easy to see that the complex roots only come in conjugate pairs, so any even degree polynomial can have all of its roots be complex. If there are an odd number of terms, the additional odd root must be real.

Basically the idea is that the only way to obtain a real result when the roots are multiplied together, the roots must be complex conjugate pairs.
 

    eecs4ever

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eecs4ever

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Re: Question about roots

jayc, your intuition is good.

heres a proof i found online. this is more rigorous.
 

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