Function with no roots Given a non-constant function $f(x)$, is it possible for it to have no zeroes (neither real nor complex)?  Say for example, $f(x)=\cos x-2$, does a complex solution exist for this because for real $x$, $\cos x$ belongs to $[-1, 1]$? Or, say $f(x)=e^x$, does a complex zero exist for this because $e^x \gt0$ for real $x$? 
 A: $\exp(x)$ has no real roots, and no complex roots either.
It is not difficult to find functions that have no roots: for any function $f(x)$, $g(x) = |f(x)|+1$ has no roots.
A: The answer to the question as asked is simply "yes", as others have said. I'd like to give a little more context and explain why (for one interpretation of the question) the answer comes close to being no.
So, first of all, "function" is a very broad term. The usual definition in mathematics is that a function is any way of assigning output values to input values, and obviously with this definition it's very easy to have a nonconstant function on $\Bbb{C}$ that's never zero; e.g., let f(z)=1 unless z=9 in which case f(z)=2.
I take it the questioner is interested in "nice" functions in some sense, and in particular I'm guessing she has in mind functions that are built in the obvious way out of "standard" functions like addition, sin, exp, etc.
There is an important notion in complex analysis, of an analytic function. That means a function that's "complex-differentiable"; that is, for any $z$ there's a complex number $a$ such that $f(z+h)=f(z)+ah+o(|h|)$, that last term denoting something that $\rightarrow0$ faster than $|h|$ does. (We then write $f'(z)=a$.)
Now, analyticity turns out to be an extremely stringent condition. For instance, if you know a function is analytic and you know its values at the numbers $1/n$ then that completely determines all its values everywhere! But because analyticity is just "complex differentiability", if you start with some analytic functions (e.g., constants, $f(z)=z$, sin, exp, ...) and combine them with addition, subtraction, multiplication and function composition -- e.g., $f(z)=\cos(\sin(2z-\exp(3z))+7z^5)-\exp(z^2)$ -- the result will still be an analytic function.
So we might want to take the original question as being specifically about analytic functions. As, e.g., lhf has said, the answer is still "yes". But now it is only just "yes". Here's why.
Picard's theorem says that if you have a non-constant analytic function from all of $\Bbb{C}$ to $\Bbb{C}$, then there is at most one value that it never takes. So, e.g., you can find such a function so that $f(z)=0$ has no solutions; but then $f(z)=w$ will have solutions for every $w\neq0$.
And lhf has pointed out another way in which the answer is only just barely "yes": the only entire functions ("entire" is shorthand for "analytic on all of $\Bbb{C}$") that never take the value $0$ are those with the rather special form $f(z)=\exp g(z)$ where $g$ is also an entire function.
A couple of words of caution. First: some "nice" functions aren't analytic. For instance, what about taking square roots? Well, even in the real numbers, if we put $f(x)=\sqrt{x}$ then $f'(0)$ is infinite. So: not analytic. Second: if a function becomes infinite anywhere (like, say, $1/z$ or $\tan z$) then it isn't analytic. (But it might be meromorphic, which roughly means it's the ratio of two analytic functions, and the appropriate version of Picard's theorem then says it's allowed to miss at most two points, one of which might be $\infty$.)
A: It is a property of polynomials that a polynomial of degree $n$ must have exactly $n$ complex roots. For example, if you took a function like:
$$ f(x) = 3x^2 + 2x -4$$
Its highest power is 2, so it will have two complex roots, and this can be said for any polynomial with real or complex coefficients. (However, some roots may have "multiplicity," so that a polynomial of degree $n$ has fewer than $n$ values which yield zero; some of these are counted as more than one root.)
However, no such theorem exists for most other functions you encounter: such as exponentials, sinusoids, logarithms and many others. They simply may just have no roots at all, or they can have multiple; we just don't have a guarantee that roots always exist like we do for polynomials.
A: The $f(x)>0$ for real $x$ condition is mostly irrelevant. 
If a continuous function $f(x)$ with no real roots is defined on the whole real line, then it necessarily must satisfy $f(x) >0$ for all real $x$ or $f(x)<0$ for all real $x$, as otherwise it would have a real root by the intermediate value theorem. However, this doesn't imply that it had no complex roots. As an example, $x^2 +1>0$ for all real $x$, but it has the complex roots $\{i,-i\}$. But $e^x$, as commenters have pointed out, is also positive for all real $x$ and has no complex roots. 
And if a function is not continuous or not defined on the whole real line, it may not satisfy that condition and yet have no complex roots. For instance, $\frac{1}{x}$ takes on both positive and negative real values, but had no roots anywhere in the complex plane (but it has a pole at 0, so the intermediate valid theorem doesn't hold).
A: The gamma function $ z \mapsto \Gamma(z)$ has no roots on the complex plane.
Observe that $$\cos \left(i\log(2+\sqrt{3})\right)=2$$ thus $\cos x -2=0$ does admit a complex root.
Edit. From $\displaystyle \cos x =\frac12\left(e^{ix}+e^{-ix}\right)$, one has
$$
\begin{align}
\cos \left(i\log(2+\sqrt{3})\right)&=\frac12\left(e^{i\left(i\log(2+\sqrt{3})\right)}+e^{-i\left(i\log(2+\sqrt{3})\right)} \right)
\\&=
\frac12\left(e^{-\log(2+\sqrt{3})}+e^{\log(2+\sqrt{3})} \right)
\\&=\frac12\left(\frac1{2+\sqrt{3}}+2+\sqrt{3}\right)
\\&=\frac12\left(\frac{2-\sqrt{3}}{4-3}+2+\sqrt{3}\right)
\\&=\frac12\left(2-\sqrt{3}+2+\sqrt{3}\right)
\\&=\color{red}{2}.
\end{align}
$$
A: If $g$ is an entire function, then $f(z)=e^{g(z)}$ is entire and has no zeros in the complex plane.
This is essentially the only example for $f$ entire because of the Weierstrass factorization theorem.
