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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?
M.Mass
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Mark Macheta
  • 23

1 Answers1

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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 $(-1)^{3/2}=e^{i3\pi/2}=-i.$

David K
  • 98,388