$(a^b)^c$ and $a^{(bc)}$ for complex numbers

Question

I've just stumbled across $$\left(e^{2\pi i}\right)^{\frac{1}{2}}=\sqrt 1=1\neq -1=e^{\pi i},$$do I have some error in my thoughts there or does $(a^b)^c=a^{(bc)}$ not hold for complex numbers?

Answer 1

Actually, that equality makes no sense for complex numbers. How do you even define $a^b$ if $a$ and $b$ are arbitrary complex numbers?

Answer 2

That's because $$ (e^{2\pi i})^{1/2}\neq e^{\pi i}, $$ even though their square is equal: $$ (e^{2\pi i})= (e^{\pi i})^2. $$ Just like what's in real numbers, every complex number have two square roots. The power $1/2$ placed at the top right corner just gives you one of them. $e^{\pi i}$ is another square root of $1$.

Answer 3

The complex numbers are just a distraction. We can rewrite your example as

$$(-1)^2=1 \implies \big ((-1)^2 \big)^{1/2}=1^{1/2} \implies-1 = \sqrt 1 = 1 \implies -1 =1.$$

Do you see where we made the mistake above?