Well, to illustrate my point the differential equation f'(x) = 1/x has the solution log(x) for positive x. If you let x become negative then that particular sol

Well, to illustrate my point the differential equation f'(x) = 1/x has the solution log(x) for positive x. If you let x become negative then that particular solution ends up becoming complex, however there is still a real solution namely log(-x) (which only differs by a constant). The Black Scholes equation is a tad too complicated to say whether something like that will happen but it's too simple to just conclude everything breaks because of one complex logarithm.