Question regarding the proof of $\pi$ is transcendental.

Question

I was reading a proof of the fact that $\pi$ is transcendental over $\mathbb{Q}$, and the author mentioned that if $\pi$ is algebraic then so is $i \pi$, but I'm not particular sure why. Can anybody explain this to me? Thank you.

Answer 1

Suppose that $i\pi$ is algebraic. Then there are rational numbers $a_0,a_1,\ldots,a_n$, not all of which are $0$, such that $i\pi$ is a root of $P(x)=a_0+a_1x+\cdots+a_nx^n$. Now, let$$Q(x)=P(ix)=a_0-a_2x^2+\cdots+(a_1x-a_3x^3+\cdots)i.$$Then $Q(\pi)=P(i\pi)=0$. So, since $\pi\in\Bbb R$, $\pi$ is a root of both polynomials$$Q_1(x)=a_0-a_2x^2+\cdots\quad\text{and}\quad Q_2(x)=a_1x-a_3x^3+\cdots,$$and at least one of them is not the null polynomial. But then $\pi$ is a root of $Q_1(x)^2+Q_2(x)^2$.