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I am new to working with complex numbers and am confused about using existing methods of working with indicies.

Consider that:

\begin{equation}\begin{aligned} & x = -1\\ &(\sqrt{x})^{3}=-i\\ \end{aligned}\end{equation}

I am perfectly comfortable with this result, given that using exponential form and raising to a power of 3/2 gives the same result.

My problem is, if you apply the cube before the square root, you get a different result:

\begin{equation}\begin{aligned} &\sqrt{(x)^{3}}=i \end{aligned}\end{equation}

\begin{equation}\begin{aligned} & \text{Since } (-1)^{3} = -1. \end{aligned}\end{equation}

So in this case:

\begin{equation}\begin{aligned} (\sqrt{x})^{3} \ne \sqrt{x^{3}} \end{aligned}\end{equation}

Why is this problem introduced and is there an explanation of how to resolve this other than saying the normal rules don't apply? I want to be sure about when I can apply certain operations when working with complex numbers and to this point any new mathematics I have learned is completely consistent with existing rules (meaning you cannot get a false result).

ajax2112
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1 Answers1

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Short answer: fractional and real exponents applied to complex numbers don't behave the way you expect. Natural-number exponents do. Integer exponents do.

In particular, you've tried to use the rule that $$ x^{\frac a b} = (x^a)^\frac1b, $$ and that one doesn't work in general. That's sort of a pity, but it's also the starting point for a lot of interesting mathematics.

All the other rules you know and love, like $$ x^{a + b}= x^a \cdot x^b $$ work fine, too. But just as, for real numbers, $\log$ is defined only on the positive real-axis, so rules like $$ \log(a^b) = b \log a $$ only make sense for positive values $a$, for the complexes $\log$ is defined on a larger domain...so things have the potential to be more subtle.

If you want to know what rules apply, one possibility is to work with the actual definitions and check.

John Hughes
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  • Thank you very much. An additional question though: I have seen the use of natural logarithms to find the value of things like i^i, allowing the exponent to be taken out of the logarithm and multiplied. Could you be a little bit more specific as to when logarithms apply? – ajax2112 Apr 20 '20 at 03:51
  • If it were 1975, when I took my first complex variables course, I could. Even if it were 1982, when I taught the subject once. But now...no. I'd have to go look stuff up and be really careful, and I've got other things I need to do. And now that you know there might be a problem, you can look them up too, and it'll do you more good than it'll do me. – John Hughes Apr 20 '20 at 11:46