Why can't the exponent laws be extended to complex numbers?

$(x^a)^b=(x^b)^a=x^{ab}$ only when real numbers are involved. This implies that the base must be a positive number. For example, with $x=-1$, $a=2$, and $b=\frac{1}{2}$, $(x^a)^b=\{(-1)^2\}^\frac{1}{2}=1$; but with $x=-1$, $a=\frac{1}{2}$, and $b=2$, $(x^a)^b=\{(-1)^\frac{1}{2}\}^2=-1$. So the rule breaks when a non-real number is involved ($(-1)^\frac{1}{2}$). The question may sound a bit weird, but why is it so? And is there a special set of exponent laws that work for the complex numbers (as well as real numbers, as a subset)?

• Think about square root on the complex plane. It can't be defined everywhere. If you try to define it everywhere continuity has to break somewhere. That's why you need branches of such functions, or to define multi-valued functions, or to define functions on Riemann surfaces. – Gregory Grant Jan 25 '16 at 20:49
• Because exponential function is periodic over the complex numbers $$e^x = e^{2\pi i n + x}$$ – Kaster Jan 25 '16 at 20:50
• @Surb Yes it is. The operation $(a,b)\mapsto (x^a)^b$, however, isn't, at least for general complex numbers. – Wojowu Feb 3 '16 at 11:03