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Assume m, b are positive integers.

$(-1)^{3m/2}$ can be written as $\sqrt{(-1)^{3m}}$. If m is even, then $(-1)^{3m}=1$, otherwise $(-1)^{3m}=-1$. Now $\sqrt{1}=1$ and $\sqrt{-1}=i$.

But why is then $(-1)^{3m/2}\cdot b^{3m/2} = (-b)^{3m/2}$?

Twistios_Player
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  • Because $a^x\cdot b^x=(a\cdot b)^x$ – Robert Feb 05 '23 at 14:57
  • @Robert And what is the reason that it is not $(-1)^{3m/2}\cdot b^{3m/2} = i \cdot b^{3m/2}$? – Twistios_Player Feb 05 '23 at 15:02
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    Fractional powers of negative numbers are not uniquely defined, and the "general rule" $(a^m)^n=a^{m\times n}$ does not always work when $m$ and $n$ are not integers; cf. this question – J. W. Tanner Feb 05 '23 at 15:05
  • Hint: compute the two sides of your expression for $m=4$ and compare them as complex numbers – JBuck Feb 05 '23 at 15:10
  • "$(-1)^{3m/2}$ can be written as $\sqrt{(-1)^{3m}}$."

    Not exactly. The symbol $\sqrt{\cdot}$ is used solely for positive square roots in the real numbers. So if $m$ is even, then $\sqrt{(-1)^{3m}} = \sqrt{1} = 1$. This does not mean $(-1)^{3m/2} = 1$, because for complex numbers, fractional powers give you multiple values. So in fact, $(-1)^{3m/2} = \pm 1$. Meanwhile, if $m$ is odd, $\sqrt{(-1)^{3m}} = \sqrt{-1} = i$, but $(-1)^{3m/2} = ((-1)^{3m})^{1/2} = (-1)^{1/2} = \pm i$.

     – D Ford Feb 05 '23 at 15:11
  • Is the rule $a^x\cdot b^x = (a\cdot b)^x$ a general rule, in the sense that it is working also for negative integers / real numbers? – Twistios_Player Feb 05 '23 at 15:25
  • @Twistios_Player That rule applies if the exponents are integers and bases are nonzero. In real analysis, it applies generally as long as the bases are positive. – ryang Feb 05 '23 at 21:50

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