is $(1+\pi)/(1-\pi)$ a transcendental number?

Question

I know it's a open problem to show if $\frac{\pi}{e}$ is a trans. number or not. But what about quotient between numbers in function only of $\pi$, which is trans, such as $\frac{1+\pi}{1-\pi}$ or $\frac{1+i\pi}{1-i\pi}$. Wolfram says all theses numbers are trans.

$\frac{1+\pi}{1-\pi} = -1- \frac{2}{\pi-1}$

I'm summing a not trans number with a trans. That implies the result is trans? Also, how to be sure that $\frac{2}{\pi-1}$ is trans? Is a division between "equal" trans numbers also trans? by "equal" I'm mean when both numbers in the division are written in function of $\pi$. (except cases like $\pi/\pi=1$)

$\frac{1+i\pi}{1-i\pi} = -1+ \frac{2i}{\pi-i}$.

Same problem.

The same question can be asked about $e$ instead of $\pi$.

Answer 1

$1-{2\over{\pi-1}}$ is algebraic implies that ${1\over{\pi-1}}=a$ where $a$ is algebraic, this implies that $\pi-1=1/a$ and $\pi=1+1/a$ contradiction

Answer 2

If $\frac{1+\pi}{1-\pi}$ were algebraic, there would be some polynomial $P(x)$ (with rational or integer coefficients) such that $P(\frac{1+\pi}{1-\pi}) = 0$. Suppose that $P$ has degree $n$; then $Q(x) = (1-x)^n P(\frac{1+x}{1-x})$ is also a polynomial (with rational or integer coefficients), because the factor of $(1-x)^n$ clears denominators. Then, $Q(\pi) = (1-\pi)^n P(\frac{1+\pi}{1-\pi}) = 0$, which would prove that $\pi$ is algebraic.

But $\pi$ is transcendental, and therefore so is $\frac{1+\pi}{1-\pi}$.