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: .
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.
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.
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. .