Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.


(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.

share|improve this question
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.

share|improve this answer
"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 $-64=(-1)\times 8^2 \rightarrow (-64)^{3/2}=(-1)^{3/2}(8^2)^{3/2}=(-1)^{3/2}8^3$

share|improve this answer
@DavidMitra you are right, typographic –  Stefanos 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... –  Stefanos Feb 28 at 13:51

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.