Robert's answer tells you correctly how to find, *given an $x$*, a complex $c$ such that $x^c=-x$.

However, it is worth pointing out that you cannot find a *single* $c$ such $x^c=-x$ will hold for *every* $x$. There's no existing number that achieves this, and if you try to extend the number system by simply postulating that $c$ exists (which I think was the point of the question), you run into this:
$$ 0 = \frac08 = \frac{-4+4}{8} = \frac{4^c+4}8 = \frac{2^c2^c+4}8 = \frac{(-2)(-2)+4}8 = \frac{4+4}8 = 1 $$
So if you want your $c$ to exist, but don't want to collapse everything into $0=1$, then you have to discard the exponentiation rule $(ab)^c=a^cb^c$. And without *that* rule, the result isn't really worthy of being called exponentiation in the first place.

I think you have been fooled by a common introduction to complex numbers, which goes like this: "There is no real number whose square is $-1$, but *assume* we had such a number anyway and call it $i$. Then we can calculate with the assumption $i^2=-1$, and, ta-dah! we have invented a new kind of numbers". This presentation, however, sweeps an important point under the carpet, namely that we had better be sure that assuming $i^2=-1$ doesn't allow us to prove falsehoods such as $0=1$ about the real numbers that we already know and love -- because if it does that means that we *haven't* invented a new kind of numbers but simply deceived ourselves.

For $i^2=-1$ it turns out that it's not too difficult to prove that it doesn't lead to nonsense. But even when this proof is shown, something is still swept under the carpet here, namely that if we assume $i^2=-1$ *and* the familiar rule that every number is either 0 or positive or negative (and the usual rules for positive and negative numbers), then we can *still* derive nonsense. So what's really true is that we can get a $\sqrt{-1}$ *if and only if* we're willing to drop what we know about positive and negative numbers (or "less than"/"greater than") when we're working with complex numbers.

Extending the concept of number is almost always connected to a loss of some rules that used to hold before the extension. How far we can push the extension depends on how many rules we're willing to let go of. So you can have your $c$, but the cost you must pay is the rules for how exponentiation works. And if you pay that cost, having $c$ becomes pretty useless anyway, so the whole expedition turns out to be pointless.

(But of course we couldn't have known that before we went and checked. The *idea* was good enough; it just didn't work. Perhaps your next one will.)