Proof that the coefficients of a polynomial are real

Question

How does one prove that all the coefficients of this polynomial: $$(x+i)^{10}+(x-i)^{10}$$ are real numbers, without using Newton's Binomial Theorem?

Answer 1

If you agree with this:

If $p$ is a complex polynomial such that: $$\forall x\in\mathbb{R},\quad p(x)\in\mathbb{R}$$ then all the coefficients of $p$ are real

you'll find a good strategy!

Answer 2

$$\overline{(x+i)^{10}+(x-i)^{10}} =\overline{x+i}^{10}+\overline{x-i}^{10}=(\overline x-i)^{10}+(\overline x+i)^{10}$$

Answer 3

Fact: A complex number $z$ is real iff $z=\overline{z}$

Fact: $z\rightarrow \overline{z}$ is a field automorphism of $\mathbb{C}$

$$\overline{(x+i)^{10}+(x-i)^{10}}=\overline{(x+i)^{10}}+\overline{(x-i)^{10}}=(\overline{x+i})^{10}+(\overline{x-i})^{10}=(x-i)^{10}+(x+i)^{10}$$

Answer 4

Here's a way to do it without relying on a general theorem about complex conjugates.

Let $P(x)=(x+i)^{10}+(x-i)^{10}$ be the polynomial in question. Let

$$Q(x)=P(ix)=(ix+i)^{10}+(ix-i)^{10}=-(x+1)^{10}-(x-1)^{10}$$

The coefficients of $Q$ are obviously real. But also, $Q(-x)=Q(x)$, so $Q$ is an even function of $x$, i.e., $Q(x)=R(x^2)$ for some polynomial $R$ with real coefficients. But this leads to

$$P(x)=Q(-ix)=R(-x^2)$$

hence the coefficients of $P$ are real (and $P$ is also an even function of $x$).