I just read about a Wikipedia page on Fundamental Theorem of Algebra, and it says

"Some proofs of the theorem only prove that any non-constant polynomial with real coefficients has some complex root. This is enough to establish the theorem in the general case because, given a non-constant polynomial $p(z)$ with complex coefficients, the polynomial $$q(z)=p(z)\overline{p(\bar z)}$$ has only real coefficients and, if z is a zero of q(z), then either z or its conjugate is a root of p(z)."

I don't understand why $q(z)$ here has only real coefficients.

Suppose $p(z)=\sum_{i=0}^na_iz^{i}\in\mathbb {C}[z].$ Then $\overline{p(\bar z)}=\sum_{j=0}^n\bar {a_j}z^j.$ (Right?) We have

$$p(z)\bar{p(\bar z)}=\left(\sum_{i=0}^na_iz^{i}\right)\left(\sum_{j=0}^n\bar {a_j}z^j\right)=\sum_{k=0}^{2n}\left(\sum_{i+j=k}a_i\bar{a_j}\right)z^k.$$ However, it seems not true that $\sum_{i+j=k}a_i\bar{a_j}$ is always a real number, for some $k$. What's wrong here?

Moreover, I was wondering whether $p(z)p(\bar z)$ has only real coefficients. Why do people like choosing $p(z)\overline{p(\bar z)}$?

Thanks in advance.


2 Answers 2


Why does it seem not true to you that $\sum_{i+j=k} a_i \overline{a_j}$ is real? Its complex conjugate is $\sum_{i+j=k} \overline{a_i} a_j$, and then you can interchange $i$ and $j$.


Take, just as an example, k=2. Then you have $a_0\overline{a_2} + a_1\overline{a_1} + a_2\overline{a_0}= 2\textrm{Re}[a_0\overline{a_2}] + |a_1|^2$, definitely a real number.


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