# In the context of the fundamental theorem of algebra, why does $p(z)\overline{p(\bar z)}$ have only real coefficients?

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 $$p(z)\overline{p(\bar z)}$$?

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.