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Pulling a negative out of a square root

  1. Jan 18, 2016 #1
    The following is invalid, since the operation is not defined when ##a, b < 0##: ##\sqrt{-1}\sqrt{-1} = \sqrt{(-1)(-1)} = \sqrt{(-1)^2} = \sqrt{1} = 1##. This is not correct, because ##ii = -1##. This shows that ##\sqrt{a}\sqrt{b} = \sqrt{ab}## is invalid when ##a, b< 0##.

    However, say we have ##\sqrt{-5}##. In order to simplify this, we do the following: ##\sqrt{-5} = \sqrt{(-1)(5)} = \sqrt{-1}\sqrt{5} = i\sqrt{5}##. Why is this a valid manipulation given the previous statement that ##\sqrt{a}\sqrt{b} = \sqrt{ab}## is invalid when ##a, b< 0##?
     
  2. jcsd
  3. Jan 18, 2016 #2

    phinds

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    You are applying a criterion where a and b are both less than zero to a situation where only one of them is less than zero. I don't see why you would expect it to apply.
     
  4. Jan 18, 2016 #3

    BvU

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    In my math book (1971 :smile:) the range of the complex square root function is ##-{1\over 2} \pi \lt \arg {\bf z} \le {1\over 2} \pi ##.

    With the range of ##\phi = \arg {\bf z} ## : ##- \pi \lt \phi \le \pi ##, and using euler ## \ {\bf z} \equiv |{\bf z}| e^{i\phi} ## we can now define $$ \ \sqrt {\bf z} \equiv\; \sqrt{|{\bf z}|} \; e^{i {\phi\over 2} } $$
    With this definition of the complex square root you can see that ##\sqrt {\bf \alpha\beta} = \sqrt {\bf \alpha} \sqrt {\bf \beta} ## only holds if ## \ |\arg {\bf \alpha} + \arg {\bf \beta} | < \pi ##

    Check out the euler formula, de moivre theorem.
     
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