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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.

William
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1 Answers1

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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 $a>0$.

Martin Argerami
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