Exercise 4.9

Question

Suppose $p \in \mathcal{P}(\mathbf{C})$. Define $q:\mathbf{C}	o \mathbf{C}$ by:$$
q(z)=p(z)\overline{p(z)}.
$$Prove that $q$ is a polynomial with real coefficients.

Moon Flower · Accepted Answer

Let $p$ be of degree $n$ and $p(z)=a(z-r_{1})\dots(z-r_{n})$, where $a,r_{1},\dots,r_{n}\in \mathbb{C}.$

Since conjugates distribute over multiplication, \overline{p(z)}=a(\overline{z-r_{1}})\dots(\overline{z-r_{n}})$.

Behold: $$q(z)=p(z)\overline{p(z)}=a^{2}(z-r_{1})(\overline{z-r_{1}})\dots(z-r_{n})(\overline{z-r_{n}}).$$Firstly, $q$ is the product of two polynomials and is therefore a polynomial.

Secondly, since a complex number times its conjugate is a real number, for all inputs $z\in C$ to $q$, we will get a real number as an output. Additionally, since all of its outputs are real, all of its coefficients are real as well (if there were coefficients in the polynomial that had $i$ in them, the fact that they cancel out for all inputs means that they are equivalent to $0$).

Thus, $p$ is a polynomial with real coefficients. $\blacksquare$

Exercise 4.9

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Exercise 4.9

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