Extension of field, $\Bbb{R}(i \pi) = \Bbb{C} $ $K$ it's a field. Let $L|K$ ($L$ is extension of $K$), $a\in L$. If $a$ is an algebraic with degree $n$ over $K$, then a set $\{1,a,a^2,...,a^{n-1}\}$ is a basis of $K(a)$ over $K$. Also $K(a) = \{a_0 + a_{1}a+a_{2}a^2+...+a_{n-1}a^{n-1}, a_0,...,a_{n-1} \in K \} $ 

Why $\Bbb{R}(i \pi) = \Bbb{C} $ ?

$i\pi$ is a root of a polynomial with coefficients from $\Bbb{R}$. But I don't see degree of $i\pi$.
 A: $\pi \in \mathbb{R}$ so $\pi^2 \in \mathbb{R}$ 
$i\pi$ is a root of
$$
x^2+\pi^2 =0
$$
A: Why is
$\Bbb R(i\pi) = \Bbb C?  \tag 1$
The easiest answer I know is:
We have
$\pi \in \Bbb R \subseteq \Bbb R(i\pi) \subseteq \Bbb C, \tag 2$
and 
$i\pi \in \Bbb R(i \pi) \subseteq \Bbb C; \tag 3$
thus both
$\pi, i\pi \in \Bbb R(i\pi); \tag 4$
since $\Bbb R(i \pi)$ is a field and $\pi \ne 0$, we have
$i = \dfrac{i \pi}{\pi} \in \Bbb R(i \pi); \tag 5$
thus, for
$z = x + iy \in \Bbb C, \; x, y \in \Bbb R, \tag 6$
it follows from (2) and (5) that
$x + iy \in \Bbb R(i \pi), \tag 7$
and we conclude that
$\Bbb C \subseteq \Bbb R(i \pi); \tag 8$
we can also argue that
$x + iy = x + (i \pi) \dfrac{y}{\pi} \in \Bbb R(i \pi). \tag 9$
What we see here is that $\Bbb C = \Bbb R(i) = \Bbb R(i \pi)$; $\{1, i \}$ and $\{1, i\pi \}$ are both bases for $\Bbb C$ over $\Bbb R$.
Note that
$[\Bbb R(i \pi): \Bbb R] = 2, \tag{10}$
since the polynomial
$x^2 + \pi^2 \in \Bbb R[x], \tag{11}$
of which $i\pi \in \Bbb R(i\pi)$ is a root, is irreducible over $\Bbb R$, which follows from the fact that any non-constant factor of a quadratic polynomial must be linear, and if
$x^2 + \pi^2 = (x - \alpha)(x - \beta), \; \alpha, \beta \in \Bbb R, \tag{12}$
then
$\alpha^2 + \pi^2 = \beta^2 + \pi^2 = 0, \tag{13}$
but no real square can equal $-\pi^2 < 0$.
The common assertion that $\pi$ is a transcendental number refers to the fact that it is transcendental over $\Bbb Q$; that is, $f(\pi) \ne 0$ if $f(x) \in \Bbb Q[x]$; indeed, both $\pi$ and $i \pi$ are transcendental over $\Bbb Q$, but algebraic over $\Bbb R$.
