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$\left(-64\right)^{\left(\frac{3}{2}\right)}$

(Disclaimer - I work in a HS math center, helping students. This is from an Algebra/Trig text used by both sophomores and juniors depending on the class. Every so often, I hit an odd situation that I don't recall how to solve. Is this a paradox, or is there a preferred answer?)

I read thru How do you compute negative numbers to fractional powers? twice, and the issue isn't with i, the students know how to handle a simple square root of -1, it's more practical - do I suggest they raise to the power first, take the root first, or object to a non-integer power of a negative number? Much of the answer linked is well beyond their level.

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The usual way is to just write it as $e^{\frac32 \log(-64)}$ and take the value on the principal branch of the complex logarithm. –  user2345215 Feb 28 at 13:11
    
If $i$ is known, the calculation in the your example is relative easy $(-64)^\frac{3}{2} = (-64)(-64)^\frac{1}{2} = -64\times 8i = -512i,\;$ but I admit that this method cannot be generalized. –  gammatester Feb 28 at 13:18
    
User - understood, but beyond my students. @gammatester - I like your approach, thanks. the other method simply got me to 512i and that was what created the issue. –  JoeTaxpayer Feb 28 at 13:31

2 Answers 2

up vote 3 down vote accepted

Personally I would square root first because the students know how to do this, giving $(8i)^3=-512i$.

As with all square root questions there are two possible answers - if you cube first and then take the square root you get $512i$. This suggests that there is a hidden convention at work - indeed there is, and it is a more subtle thing when complex numbers are involved.

Personally, I don't see the point of asking a question like this, which really belongs with Argand Diagrams and the modulus/argument form of complex numbers at the elementary level. It skates over too many issues which are liable to confuse without engaging the mathematical equipment or language for discussing them.

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"I don't see the point of asking a question like this." - I agree. I'm only in this position a few months now, and finding my issues are more often than not, with the texts used. This is one by Paul Foerster, who I found to be a pretty decent author. Surprised when this came to my attention. –  JoeTaxpayer Feb 28 at 14:57

Maybe you can say that $-64=(-1)\times 8^2 \rightarrow (-64)^{3/2}=(-1)^{3/2}(8^2)^{3/2}=(-1)^{3/2}8^3$

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@DavidMitra you are right, typographic –  Stef Feb 28 at 13:01
    
(-1)^{3/2} is then the same issue, no? –  JoeTaxpayer Feb 28 at 13:32
    
yes it is the same issue...but they can have an algorithm with dealing with $(-1)^k$ like gammaster and deal with all the other numbers like my suggestion... –  Stef Feb 28 at 13:51

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