Polynomial with Root $\pi+ei$

Question

How do I find a polynomial with a root of $\pi +ei$ in the reals?

I know how to do this with algebraic numbers, but not transcendental ones like e and $\pi$.

Edit:

I now realize that in the reals, e and π are not actually considered transcendental, but I'm still a bit confused on how to proceed.

Thank you in advance!

Answer 1

If one root is $\pi +e\mathrm i$, then another root must be $\pi-e\mathrm i$.

$$\begin{align} x &=\pi \pm e\mathrm i\\ x-\pi &=\pm e\mathrm i\\ (x-\pi)^2 &=-e^2\\ x^2-2\pi x+\pi^2+e^2 &=0\\ \end{align}$$

Therefore, the polynomial $f(x)=x^2-2\pi x+\pi^2+e^2$ has a root of $\pi+e\mathrm i$.