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B Square root of a negative number in a complex field

  1. Oct 16, 2017 #1
    Mod note: Fixed all of the radicals. The expressions inside the radical need to be surrounded with braces -- { }
    (This question is probably asked a lot but I could not find it so I'll just ask it myself.)
    Does the square root of negative numbers exist in the complex field? In other words is ##\sqrt{(-a)}=\sqrt
    {(a)}*i## or is it undefined? Recently in a forum I saw a guy claiming that the square root of -10 is undefined and his proof was something like this:

    If ##\sqrt{-100}=10i## then ##\sqrt{-100}*\sqrt{-100}=10i*10i=-100##, but
    ##\sqrt{-100}*\sqrt{-100}=\sqrt{-100*-100}=\sqrt{10,000}=\sqrt{100*100}=\sqrt{100}*\sqrt{100}=100##
    Which implies ##100=-100##, a paradox. And so ##\sqrt{-100}## is not ##10i## but is instead undefined.

    I feel like he is making some sort of assumption here that I am not seeing because the square root of negatives is undefined in the complex field that kind of defeats the point of having the complex field in the first place, so is he right?
     
    Last edited by a moderator: Oct 17, 2017
  2. jcsd
  3. Oct 16, 2017 #2

    fresh_42

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  4. Oct 16, 2017 #3
    This helps, thanks :)
     
  5. Oct 18, 2017 #4

    WWGD

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    You may want to consider that ##sqrt{100}=\pm 10 ## and not necessarily 10. Similar for other roots. Depending on your choice of branch ( given that the root is defined in terms of the log) , you may have problems " jumping branches" .
     
  6. Oct 18, 2017 #5

    Mark44

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    Are you interpreting 100 as a complex number? If not, ##\sqrt{100}## is just plain old 10.
     
  7. Oct 18, 2017 #6

    WWGD

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    Yes, I sort of thought after I defined ##\sqrt## as I did, I would have to duck and cover. It is a pesky technical issue.
     
  8. Nov 4, 2017 #7
    Last edited: Nov 4, 2017
  9. Nov 4, 2017 #8

    Mark44

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    No, it has nothing to do with being an angle. See the explanation for Example C a little way down the page.
     
  10. Nov 4, 2017 #9
    That explanation depends on the 2 pi being an angle, and as far as I can remember, so does any proof of Euler's.

    Is not the 2 pi the angle of the polar co-ordinate of a unit length position vector?
     
    Last edited: Nov 4, 2017
  11. Nov 4, 2017 #10

    fresh_42

    Staff: Mentor

    The point is, that the exponential (logarithm) function behaves differently as real or complex function.
    The complex exponential function is periodic with the complex period ##2\pi \mathrm{i}##, i.e. not bijective anymore. ##\displaystyle \exp (z + 2 \pi k \mathrm{i}) = \exp (z), \; k \in \mathbb {Z}.## By restriction of the domain to a strip ##\displaystyle \{z \in \mathbb{C} \, | \, a <\operatorname{im} (z) < a + 2 \pi \}## with ##\displaystyle a \in \mathbb{R}## it has a well-defined inverse, the complex logarithm.

    This means that after a full circle, the function values are repeated and the inverse function cannot be taken anymore. So either we deal with infinitely many points all mapped to the same function value, or we restrict ourselves to angles strictly less than a full circle. Depending on the situation, both may be suitable to do. My teacher used to compare it with a spiral cut of a radish: Imagine you cut the complex number plane along the positive real axis. Then every full circle leads you to another next branch like in the picture here (just stacked):
    daikon3upload.jpg
    Source: http://photos1.blogger.com/blogger/1879/2473/1600/daikon3upload.jpg
     
  12. Nov 4, 2017 #11

    FactChecker

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    Not only is the square root of negative numbers defined in complex numbers, but many would consider that the main reason for using complex numbers.

    The problem your friend points out can also be a problem in the square root of a positive number, since 10*10 = 100 and also -10*(-10) = 100. So it is just necessary to be careful with the square root function which is multiply defined.
     
  13. Nov 4, 2017 #12
    yes, but this is because you are dealing with a 2d system rather than 1d, and all the error mentioned are due to considering the 2d system as 1d, when in fact the 1d system is just a special case of the 2d system, in particular, the fact that a complex number is a 2d position vector. the Euler equation only works taking this into account, that you are representing the vector in polar co-ordinates, but I could just as easily use degrees, or %age of full circle, or whatever angular measurement I choose to define. The point being: the fact that x is an angle is everything, not, as you say, nothing, and that is clearly the mistake being made in C. Since while 2 x 3.14... is not = 0, if we recognise that x is the angle of a unit vector polar co-ordinate in radians, as we should when using Euler's equation, then 2 x pi = 0 is correct.
     
  14. Nov 4, 2017 #13

    Mark44

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    Your argument doesn't hold water. An angle of ##2\pi## (or 360°) is not the same as an angle of 0 (or 0°), even though the positions of the terminal points of the rays that define each angle are equal. What you are saying is that there is no difference between one complete rotation and no rotations at all.
     
  15. Nov 4, 2017 #14
    True, it is not the same angle, but the vectors (ie the complex numbers) with those angles are the same, that is my point.
     
  16. Nov 4, 2017 #15

    Mark44

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    But by convention, the expression ##\sqrt{100} = 10##, the principal square root of 100. For this reason it is erroneous to write ##\sqrt{100} = \pm 10##.
     
  17. Nov 4, 2017 #16

    fresh_42

    Staff: Mentor

    In this case you placed a projection between the origin of your vector and the image after ##n## full circles. They are not the same thing, they are preimage and image. You must not arbitrarily identify them - there is a transformation in between!
     
  18. Nov 4, 2017 #17
    so 2 x 1 = 2, this isn't the same 2, therefore 2 != 2 ?
     
  19. Nov 4, 2017 #18

    Mark44

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    It is the same 2, since 1 is the multiplicative identity in the field of real number. IOW, if you multiply any real number by 1, you get back the same number.
    But that's completely differernt from saying that ##2\pi = 0##, which is obviously (I hope!) not true.
     
  20. Nov 4, 2017 #19

    FactChecker

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    That is true, but I assume that the OP did not intend that type of distinction in the context of his question. The logic that his friend was using did not recognize the principle branch or any convention for the √ symbol.
     
  21. Nov 4, 2017 #20

    Mark44

    Staff: Mentor

    I get that, but my comment had to do with a couple of posts that deal strictly with the square root of a positive real number.

    ... but not for positive real numbers.​
     
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