We know that $\mathbb C$ is an algebraically closed field with characteristic $0$.
It seems that if a proposition that can be expressed in the language of first-order logic is true for an algebraically closed field with characteristic $0$, then it is true for $\mathbb C$ (and for every algebraically closed field with characteristic $0$).
I am looking for interesting results that would be true over $\mathbb C$ but not true over some algebraically closed field with characteristic $0$.
I am more interested in "applied" results than results in field theory.
Additional comment: The initial question is vague (but on purpose). Here is what triggered that question in my mind. Sudbery proved the following result: Let $f(z)$ be a polynomial of degree $N$ with complex coefficients, and let $f^{(r)}(z)$ denote the $r$th derivative of $f(z)$. If $f(z)$ has two or more distinct roots, then $\prod_{i=0}^{N}f^{(r)}(z)$ has at least $N+1$ distinct roots. This result does not seem to generalize easily to algebraically closed fields with characteristic $0$.
Best Answer
If you want statements that are field-theoretic in nature, then a useful contrast is between the algebraically closed field $\overline{\mathbb Q}$ (the algebraic closure of $\mathbb Q$ in $\mathbb C)$ and $\mathbb C$. The former is the smallest algebraically closed field of char. zero, while $\mathbb C$ is huge in comparison.
For example, $\mathbb C$ contains a sequence of elements $x_i$ which are all mutually algebraically independent (in other words, it contains a copy of the field $\mathbb Q(x_1,\ldots,x_n,\ldots)$ of rational functions over $\mathbb Q$ in a countable number of variables), while $\overline{\mathbb Q}$ does not contain even one element that is algebraically independent of $\mathbb Q$.
Added: Note that the above property makes it much easier to prove the Nullstellenstatz over $\mathbb C$ then over $\overline{\mathbb Q}$, for example.