Q:

How can we prove that the field of complex number is not isomorphic to the field of real numbers? 

I want to prove this using contradiction, supposing that there is a ring isomorphism between the two and then finding a contradiction. 

A:

Is there an element $a\in\mathbb{R}$ such that $a^2=-1$? 

Added from comments. 
Suppose $f\colon\mathbb{C}\to\mathbb{R}$ is an isomorphism.  It's usually required that field homomorphisms map $1$ into $1$, but it's not necessary to make this assumption here.  Indeed, set $x=f(1)$. Then
$$
x=f(1)=f(1^2)=f(1)f(1)=x^2
$$
so $x-x^2=0$, which means $x=0$ or $x=1$ because we're in a field.  If $f(1)=0$, then, for all $z\in\mathbb{C}$, we have
$$
f(z)=f(z\cdot1)=f(z)f(1)=f(z)\cdot0=0
$$
which is impossible, because $f$ is bijective (injectivity is sufficient). 
Thus $f(1)=1$, hence $f(-1)=-1$. Now $a=f(i)$ would have the property that $a^2=f(i^2)=f(-1)=-1$ and there's no such element $a\in\mathbb{R}$.

