# Naked singularity of a charged black hole

Consider the Reissner-Nordstrom metric for a black hole:

$$ds^{2} = - f(r)dt^{2} + \frac{dr^{2}}{f(r)} + r^{2}d\Omega_{2}^{2},$$

where

$$f(r) = 1-\frac{2M}{r}+\frac{Q^{2}}{r^{2}}.$$

We can write

$$f(r) = \frac{1}{r^{2}}(r-r_{+})(r-r_{-}), \qquad r_{\pm} = M \pm \sqrt{M^{2}-Q^{2}}.$$

Then $r_{+}$ is called the event horizon and $r_{-}$ is called the Cauchy horizon.

There is a curvature singularity at $r = 0$.

If $|Q|>M$, then $r_{+} < 0$, so the curvature singularity is not hidden behind the horizon.

I do not understand the final sentence.

Firstly, for $|Q|>M$, I find that $r_{\pm}$ is imaginary.

Secondly, even if $r_{+} < 0$, this does not make sense: $r$ is a radial coordinate. How can it be less than $0$?

It can't. You are right, it is simply imaginary, i.e. It does not exist. And neither does $r_-$. It means there is no horizon, but you still have the singularity. So, if it existed as a physical case, it would be a naked singularity.