Powers of $i$ in Complex Numbers. Since $i = \sqrt{-1}, i^2 = -1, i^3=-i, i^4 =1$ I understand to calculate, say,  $i^{999}$ I just have to $i^{999} = i^{4 \cdot 249 + 3} = (i^4)^{249} \cdot i^{3} = -i$
But I have a question here, why can't I do something like $i^{999} = (i^4)^{\frac{999}{4}} = 1$ clearly the answer is wrong, infact everytime I'll only end up with $1$. Can anyone point out which step exactly is invalid and with reasons like why is it wrong. 
Which naturally extends the question to, how do you calculate $i$ to the power of any "rational" exponent. Like $i^{743/5}$ or something.
Thanks.
 A: You don't need complex numbers to run into that "problem": you could have written
$$
-1=(-1)^3=[(-1)^2]^{3/2}=1.
$$
You are happily using the rule $(a^b)^c=a^{(bc)}$. To think of the scope of of this rule, you need to think about what $a^b$ means. Unless you are dealing with integer powers, or with rational powers of positive integers, things are obvious. What we usually do is to define
$$
a^b=e^{b\log a}. 
$$
This requires that you have $\log a$ defined, which is not a trivial thing when $a$ is not positive. It can be done to a certain extent (you need to think in terms of branches of the complex logarithm), but it has issues, and this is one of them. Your rule depends on 
$$
(a^b)^c=e^{c\log a^b}=e^{bc\log a}.
$$
So what you need is the rule $\log (a^b)=b\log a$. And already with negative numbers you run into problems with this: you want 
$$
0=\log[(-1)^2]=2\log(-1),
$$
so you need to define $\log(-1)=0$. But now you would get 
$$
e^{\log(-1)}=e^0=1,
$$
and now the logarithm is not the inverse of the exponential!
In summary, the rules $\log(ab)=\log a+\log b$, $\log (a^b)=b\log a$, and the associated power rules, require $a>0$. 
