1)Let be any polynomial in (complex numbers). Let be the conjugate polynomial of . For example, say then (just replace by ). Prove that is a polynomial in (real coefficients).

2)Find the infinite product: .

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- November 27th 2007, 08:56 AMThePerfectHackerProblem 42
1)Let be any polynomial in (complex numbers). Let be the conjugate polynomial of . For example, say then (just replace by ). Prove that is a polynomial in (real coefficients).

2)Find the infinite product: . - November 27th 2007, 02:49 PMmath sucks
1) Proceed by induction on the degree of . The statement is obviously true for degree 0. Now assume it is true for degree . Let be a polynomial of degree . Then it can be written as

for some polynomial with degree (or less). Similarly, let

where is the conjugate polynomial of . Now,

By induction, has real coefficients. Next, by the multiplicative property of conjugation, is still the conjugate of . Thus, when added, the imaginary parts cancel out. Lastly, is obviously real. Thus, must also have real coefficients.

2) If for some integer and odd integer , then the infinite product certainly does not converge as for all . Now if for some , then is 0, so the product diverges. Finally, if for , then the product converges to 1 if is even and -1 if is odd. - December 3rd 2007, 07:37 PMThePerfectHacker
1)Lemma: If is a zero of then is a zero of . The proof is really simple say then . Now take the complex conjugate of both sides, but the complex conjugate preserves sums and products thus . Thus, is a zero of . Now consider the product . Now if is a root of this polynomial then is a zero of WLOG, thus is a zero of and so is a zero of . We have show that any zero of also has its complex cunjugate as a zero. Thus, is a polynomial with real coeffcients. (This problem was an excercise problem in an abstract algebra book).

2)This is an infinite series that I have seen several times before. Suppuse we want to simplifiy . Multiply both sides by thus . Thus, . Thus, . Thus, . Then supposing that (that is a special case) we have . So in general if is the -th partial product then if . Now it is easy to take the limit. - June 17th 2009, 03:01 PMBruno J.
Here is another simple proof of the nice theorem

**(1)**:

Denote the complex conjugate of . Then for any two functions we have .

This follows directly from the similar property of conjugation for elements of and from the definition of a polynomial product.

Then = .

Hence is its own conjugate, i.e. .