Problems with the proof of the Transcendence of $\pi$

Question

I'm studying the Niven's proof of transcendence of $\pi$, but I can't understand a passage which seems obvious to anybody. So, we suppose $\pi$ to be algebraic, and we define $\theta_1=i\pi$ which is algebraic too.

By the definition of algebraic number, there exists a polynomial $p$ with integer coefficients such that $p(\theta_1)=0$. We may assume $p$ to be the minimal polynomial. We define $\theta_i$, for $i \in 1,\dots,d$ the conjugates of $\theta_1$, and we write $p$ as $$\star \hspace{1 cm} c_d(x-\theta_1)(x-\theta_2)\ldots(x-\theta_d)=0 $$ ($c_d$ is the leading coefficient of $p$). And here there's the problem: we all know that $e^{i\pi}+1=0$, so $e^{\theta_1}+1=0$, but I can't understand why we can write $\star$ as $$(e^{\theta_1}+1)(e^{\theta_2}+1)...(e^{\theta_d}+1)=0$$

Is there anyone who can help me? any hint or suggestion would be really appreciated! Thanks!

Answer 1

If the factor $e^{\theta_1}+1$ is $0$, then the product is $0$.

Answer 2

There are some things wrong with with what you write. First, the statement

"We may assume $p$ to be the minimal polynomial. We define $θ_i$, for $i∈1,…,d$ the conjugates of $θ_1$, and we write $p$ as $$⋆\quad c_d(x−θ_1)(x−θ_2)…(x−θ_d)=0"$$is wrong.

(1) $\quad p$ here should be $p(x)$.

(2) $\quad$The equation $\star$ is not $p(x)$: rather it is the equation of $p(x)$ to zero. This is equivalent to the assertion that $x$ is one of $\theta_1,...,\theta_d$.

The next two lines are OK. Also the displayed equation you write is true, because the first factor on the LHS is zero. However, it is not any way of writing $p(x)$; nor does it say anything about the values of $x$ that make $p(x)$ equal to zero.