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I started with

$$x=(-16)^{\frac{1}{2}}$$

$$x=(-16)^{\frac{2}{4}}$$

Since $$(a^m)^n=a^{mn}$$ we have:

$$x=((-16)^2)^{\frac{1}{4}}$$

$$x=((16^2)^{\frac{1}{4}}$$

$$x=\sqrt{16}=4$$

Hence $$(-16)^{\frac{1}{2}}=4i=4$$

$$i=1$$

nonuser
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Ekaveera Gouribhatla
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    If $m$ or $n$ are not integers and $a<0$ the formula $a^{mn}=(a^m)^n$ is false. In fact, for $a<0$ and $r\in \Bbb Q$ the meaning of $a^r$ is undefined. – ajotatxe Mar 14 '19 at 15:24

3 Answers3

6

The rule $(a^b)^c = a^{bc}$ does not hold in general except in special situations such as:

  • $a$ is a positive real and $b,c$ are real, or
  • $a$ is arbitrary and $b,c$ are integers.

For example: $((-1)^2)^{1/2} = 1$, but $(-1)^{2\cdot 1/2}=-1$.

Or: $(e^{2\pi i})^{i} = 1^i = 1$, but $e^{2\pi i\cdot i} = e^{-2\pi} \approx 0.002$.

hmakholm left over Monica
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2

The imaginary numbers invalidate some of the usual rules on the exponents.

In the first place,

$$i^2<0.$$

0

For positives a and b, $\sqrt{-a} \sqrt{-b} \not = \sqrt{ab} $ But, $\sqrt{-a} \sqrt{-b} = - \sqrt{ab} $

Aditya
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