Exponential Properties

Question

Here are my steps:

$e^{{2\pi i}/100} = (e^{\pi i})^{{2/100}} = ((-1)^2)^{1/100} = 1^{1/100} = 1$.

I'm not sure if the normal rules of exponents apply like this if the power is complex.

Answer 1

The standard rule $(a^b)^c$ does not always apply for complex numbers. For example, one may take $(e^{2\pi i})^i = (\cos(2\pi) + i sin(2\pi))^i = 1^i = 1$ or $(e^{2\pi i})^i = e^{-2\pi} \neq 1$. For complex numbers, $(e^a)^b$ = $e^{ab}$ if $b$ is an integer or $a$ is real. $1/100$ is not an integer, so it is not valid to split up the exponents in such a way. The solution would be $$e^{2\pi i / 100} = \cos(\frac{2\pi}{100}) + i \sin(\frac{2\pi}{100})$$

Answer 2

No, the usual properties of real exponentials do not all apply here. The following simple example will convince you: on the one hand, $(-1)^{\frac 2 2} = \big( (-1)^2 \big) ^{\frac 1 2} = 1 ^{\frac 1 2} = 1$; on the other hand, $(-1)^{\frac 2 2} = \big( (-1) ^{\frac 1 2} \big) ^2 = \Bbb i ^2 = -1$. This shows that you must be careful with non-integer powers of complex non-positive numbers. On the other hand, if you work only with integer powers then everything is fine.