We know that $\mathbb{C}^3$ = {(z1, z2, z3) | z1, z2, z3 $\in$ $\mathbb{C}$}
How can I show that $\mathbb{C}^3$ is a vector space over $\mathbb{R}$?
Edit: Why was this closed? it is not a duplicate, the question is asking HOW to show something, the possible duplicate does not show how to do what this question is asking.
Hint: for $z\in\mathbb{C}$, write $z=x+iy$. Define scalar multiplication by $rz=rx+iry$ and addition by $(x+iy)+(x'+iy'):=(x+x')+i(y+y')$. You can even prove that $\mathbb{C}^3\cong\mathbb{R}^6$
Edit (Since OP seems confused about this).
We have $\mathbb{C}^3=\{(z_1,z_2,z_3): z_1,z_2,z_3\in\mathbb{C}\}$. We define scalar multiplication by $r\cdot(z_1,z_2,z_3)=(rz_1,rz_2,rz_3)$ for all $r\in\mathbb{R}$ and addition by $(z_1,z_2,z_3)+(w_1,w_2,w_3):=(z_1+w_1,z_2+w_2,z_3+w_3)$.
To show that $\mathbb{C}^3$ is a vector space over $\mathbb{R}$, we need to see that
(1) $\mathbb{C}^3$ with the addition defined above is an abelian group. This is trivially true (since you asked why it is closed under addition: well, obviously $(z_1+w_1,z_2+w_2,z_3+w_3)$ is a triplet of complex numbers, that is an element of $\mathbb{C}^3$!)
(2)$ (r_1r_2)\cdot(z_1,z_2,z_3)=r_1\cdot(r_2\cdot(z_1,z_2,z_3))$ for all $r_1,r_2\in\mathbb{R}$ and $(z_1,z_2,z_3)\in\mathbb{C}^3$. Again, this is trivial.
(3) $(r_1+r_2)\cdot(z_1,z_2,z_3)= r_1\cdot(z_1,z_2,z_3)+r_2(z_1,z_2,z_3)$. Again, this is trivial.
(4) $r\cdot((z_1,z_2,z_3)+(w_1,w_2,w_3))=r\cdot(z_1,z_2,z_3)+r\cdot(w_1,w_2,w_3)$. Again, this is trivial.
(5) $1\cdot(z_1,z_2,z_3)=(z_1,z_2,z_3)$. Again... this is trivial!
If you convince yourself that (1)-(5) are true, you got it.
Now if you want to see that $\mathbb{C}^3\cong\mathbb{R}^6$, write $(z_1,z_2,z_3)$ uniquely as $(x_1+iy_1,x_2+iy_2,x_3+iy_3)$ with $x_i,y_i\in\mathbb{R}$ and define $f:\mathbb{C}^3\to\mathbb{R}^6$ by $$f(x_1+iy_1,x_2+iy_2,x_3+iy_3)=(x_1,y_1,x_2,y_2,x_3,y_3).$$ I leave it as an exercise to show that this is a linear isomorphism.