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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.

Community
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JTP - Apologise to Monica
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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 '14 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 '14 at 13:18
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    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. – JTP - Apologise to Monica Feb 28 '14 at 13:31

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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.

Mark Bennet
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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. – JTP - Apologise to Monica Feb 28 '14 at 14:57
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Apply the principles of my answer.

https://math.stackexchange.com/a/4860772/928654

Let $e$ be the base of the natural logarithm, let $i^2=-1$, let $z$ be an integer, and let $H$ be the angular measure of a half-circle.

$$(-64)^{3/2}=(e^{6\ln(2)+iπ+i2πz})^{3/2}=(e^{6\ln(2)+iπ(2z+1)})^{3/2}=e^{9\ln(2)+i3π(2z+1)/2}=512\cos(3H(2z+1)/2)+i512\sin(3H(2z+1)/2)$$

$3(2z+1)/2$ is never an integer for any integer $z$, so the imaginary part never vanishes, because the only zeroes of the sine function are integer multiples of a half-circle. However, the denominator of the exponent is 2. This means that $3(2z+1)$ always being odd, the half-circle is multiplied by half of an odd number. In other words, it is an odd multiple of a right angle. The sine of an odd multiple of a right angle is either $1$ or $-1$. As such, the real part is $0$, as the cosine of an odd multiple of a right angle is always $0$.

Applying the formula above the ($\frac{3}{2}$)th powers of $-64$ are $i512$ and $-i512$.

Michael Ejercito
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  • Thank you. Your answer reminds me of why when a student walks in with a classic 'gravity' problem I need to ask what year they are. Why does that matter? A senior taking calculus needs to find the first derivative, the sophomore? The vervex of the parabola. Assuming the simple "how high did it go?" Your answer taught me something useful for precalc students, but not the kids I was helping with this question. +1 and thx. – JTP - Apologise to Monica Feb 15 '24 at 13:03