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Obviously something is wrong with this, but where is the error and why is it one?

$$ \begin{align} \sqrt{-1} &= (-1)^{1/2} \\ &= (-1)^{2/4} \\ &= \sqrt[4]{(-1)^2} \\ &= \sqrt[4]{1} \\ &= 1. \end{align} $$

iadvd
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dsaxton
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  • The way you are using the laws of indeces is only valid for positive numbers. – MGA Jun 23 '15 at 22:55
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    Your last step is incorrect. The fourth root of $1$ has in fact four solutions in the complex plane: $+1$, $+i$, $-1$ and $-i$. Your initial value is indeed one of them. – M. Wind Jun 23 '15 at 22:58
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    This is because $\sqrt{-1}$ is imaginary, It is not there !! :D

    An illusion !

     – alkabary Jun 23 '15 at 23:00
  • With the same (wrong) reasoning, $-1=\sqrt[3]{-1}=(-1)^{1/3}=(-1)^{2/6}=\sqrt[6]{(-1)^2}=\sqrt[6]{1}=1$ – egreg Jun 23 '15 at 23:17

3 Answers3

4

Multivaluedness has been ignored at several steps, example $\sqrt{-1}=\pm i$.

In the language of complex analysis, you have to define an appropriate branch of the square root function.

felasfa
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  • +1 correct, but it may be better to include a more basic explanation. People who understand this answer don't need it :-D – Ant Jun 23 '15 at 23:02
  • @Ant, I agree. will do so next time, thanks for the input – felasfa Jun 23 '15 at 23:06
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When the exponent is not an integer there may be more than one valid result of any exponentiation of a complex number.   In particular there are always two square roots, and four quaternary roots.   The square root of a square of a number is not necessarily that number.

${(a^2)}^{1/2} = \pm a$

$(r^2e^{2i\theta})^{1/2} = \lvert r\rvert e^{i(\theta+n\pi)} \;\mathbf 1_{n\in\Bbb Z}$

Graham Kemp
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Because there is more than one root of $1^{\frac{1}{4}}$.

Aleksandar
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