How to prove that complex numbers (C) are linear space over real numbers (R) field?

Verify that it satisfies the axioms, of course!

1. The sum of two complex numbers is a complex number.
2. Multiplying a complex number by a real number yields a complex number.
3. Addition of complex numbers is associative.
4. Addition of complex numbers is commutative.
5. The complex number $0$ is such that $z+0 = z$ for all complex numbers $z$.
6. Given a complex number $z$, there is a complex number (namely, $-z$) such that $z+(-z) = 0$.
7. Multiplication by reals numbers is associative: $\alpha(\beta z) = (\alpha\beta)z$ for all $\alpha,\beta\in\mathbb{R}$, $z\in \mathbb{C}$.
8. Multiplication by reals distributes over the sum of complex numbers: $\alpha(z+w) = \alpha z + \alpha w$ for all $\alpha\in\mathbb{R}$, $z,w\in\mathbb{C}$.
9. Multiplication by a complex number distributes over the sum of real numbers: $(\alpha+\beta)z = \alpha z + \beta z$ for all $\alpha,\beta\in\mathbb{R}$, $z\in\mathbb{C}$.
10. Multiplication by the real number $1$ is the identity mapping: $1z = z$ for all $z\in\mathbb{C}$.

If all 10 are true, then $\mathbb{C}$ is a vector space/linear space over $\mathbb{R}$.

(More generally, if $\mathbf{F}$ is a field, and $\mathbf{K}$ is a field that contains $\mathbf{F}$, then $\mathbf{K}$ is a vector space over $\mathbf{F}$).

The term "linear space" is another word for vector space. There are two aspects to showing that something (call it $V$) is a vector space over a field $K$ (in this case, $K=\mathbb{R}$, the real numbers):

First, you need to specify a vector addition operation "$\star$", that takes any two elements of $V$ and produces another one, and a scalar multiplication operation "$\circ$", that takes an element of $K$ and an element of $V$ and produces an element of $V$. Second, you need to show that those operations satisfy the axioms for being a vector space.

Here, the set $V$ under consideration is $\mathbb{C}$, the complex numbers. The vector addition is that is implied is just the usual addition of complex numbers; that is, we define $$a\star b:=a+b\text{ for all }a,b\in\mathbb{C}.$$ The scalar multiplication (i.e., multiplying a vector (an element of $V$) by a scalar (an element of $K$)) that is implied is just the multiplication of a complex number by a real number; that is, we define $$\lambda\circ a:=\lambda a\text{ for all }\lambda\in\mathbb{R}\text{ and }a\in\mathbb{C}.$$

The axioms for being a vector space are here. There is not really anything substantial you need to show. For example, $1\circ a=a$ for any $a\in\mathbb{C}$ because, by definition, $$1\circ a=1a=a$$ (because $1\in\mathbb{R}\subset\mathbb{C}$ is the multiplicative identity of $\mathbb{C}$'s usual multiplication). Another example: $$\lambda\circ (a\star b)=\lambda(a+b)=\lambda a +\lambda b=(\lambda\circ a)\star(\lambda\circ b)$$ because multiplication of complex numbers distributes over addition.

$\mathbb{C} = \mathbb{R}(i) = \{a+ib:a,b \in \mathbb{R} \}$ ($\mathbb{R}(i)$ is the smallest field which contains $\mathbb{R}$ and $i$). You can see that it is a 2-dimention vector space on $\mathbb{R}$ ( for example, you can verify that $\{1,i\}$ is a basis for it).

I hope I've helped you.