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This question already has an answer here:

I was recently looking at why one of my answers turned out incorrectly using this Power Rule $(a^n)^m = a^{n\times m}$ .

Trying to solve $(-1)^{3/2} = ?$

I found that

  • $(-1)^{1/2\cdot 3}$ would become $((-1)^{1/2})^3$. This becomes $-i$.

and

  • $(-1)^{3*1/2}$ would become $((-1)^3)^{1/2}$. This becomes $i$.

I know how to solve both, and understand the answer should be $-i$.


Why can I multiply as $1/2 \cdot 3$ and not $3\cdot 1/2$ ? Are there additional conditions in the Power Rule that I have missed?

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marked as duplicate by GEdgar, Ken Duna, Trevor Gunn, user91500, Hans Lundmark Jul 9 '17 at 15:13

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ the power rule doesn't work in $$\mathbb{C}$$ $\endgroup$ – Dr. Sonnhard Graubner Jul 9 '17 at 12:36
  • $\begingroup$ $((-1)^{1/2})^3=(\{-i,i\})^3=\{-i,i\}$ and $((-1)^3)^{1/2}=(\{-1\})^{1/2}=\{-i,i\}$. It works! :p $\endgroup$ – Olivia Jul 9 '17 at 12:38
  • $\begingroup$ @Arjang Someone needs to review their complex numbers basic arithmetic. $\endgroup$ – Olivia Jul 9 '17 at 12:45
  • $\begingroup$ mathworld.wolfram.com/ComplexExponentiation.html $\endgroup$ – user451844 Jul 9 '17 at 12:56
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A more reliable rule is $$a^b=e^{b\ln(a)}.$$ Now the answer you get depends on how you interpret $\ln(a).$

If you consider the complex logarithm to be a multivalued function, you may get multiple answers. In the case $a=-1,$ $\ln(-1)=i\pi(1+2k)$ where $k$ is an integer, $\frac32\ln(-1)=i\pi\left(\frac32+3k\right)$ where $k$ is an integer, and $e^{i\pi((3/2)+3k)}$ is $i$ when $k$ is odd and $-i$ when $k$ is even. So $i$ and $-i$ both are values of $(-1)^{3/2}.$

On the other hand, if you take a branch of the logarithm in which $\ln(-1)=i\pi,$ then $(-1)^{3/2}=e^{i3\pi/2}=-i.$

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