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Taking Square Roots

  1. Apr 12, 2006 #1


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    Okay, before I start my discussion, note that [itex] k \in \mathbb{R} [/itex]

    Edit: for anyone reading this right now, I accidentally hit 'submit post' instead of 'preview', so please ignore the thread until it has some content!!!

    Okay, here is the content. In my prof's notes, he is evaluating:

    [tex] \sqrt{k^2} [/tex]

    and he writes:

    [tex] \sqrt{k^2} = |k| [/tex]

    I can understand that, because the square root sign denotes the positive square root, so in order to ensure the result is positive without making any assumptions about the intrinsic sign of k, you'd have to take its absolute value. I understand it, but it still caught me off guard!!! Maybe I'm just forgetting basic stuff, but I usually blindly write:

    [tex] \sqrt{k^2} = k [/tex]

    without even thinking about it. So here I am in university, taking complex analysis, and being confused by a stupid little thing like this. In this case, it's causing me even more confusion because it doesn't seem to immediately make a difference. He's actually solving a differential equation, and in this step he is solving the characteristic equation for a variable r using the quadratic formula to obtain this result:

    [tex] r = -1 \pm \sqrt{-k^2} = -1 \pm i\sqrt{k^2} = -1 \pm i|k| [/tex]

    That is pretty much what he has written. But the thing is, why should I even care about it? We're not just taking the positive square root. The quadratic formula accounts for both positive and negative square roots. To me, writing this:

    [tex] r = -1 \pm i|k| [/tex]

    seems silly because there is already a plus or minus, and the expression evaluates to:

    [tex] r = -1 \pm ik [/tex] if k > 0

    [tex] r = -1 \mp ik [/tex] if k < 0

    so logically this covers all the bases no matter the sign of k! Regardless of whether k equals +3 or -3, I will end up with r = -1 +/- 3i

    So...why can't I just write the result as:

    [tex] r = -1 \pm ik [/tex]

    and be done with it?

    YET, it DOES make a difference to the solution of the D.E. later on whether you write k or |k|. So, strictly speaking, is |k| the ONLY correct answer to [itex] \sqrt{k^2} [/itex] when [itex] -\infty < k < \infty [/itex]?
    Last edited: Apr 12, 2006
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  3. Apr 12, 2006 #2


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    Something just occured to me: one possible simple answer to my question is YES, you have to write the answer as |k|, because square root of k squared is a specific case of a more general formula for a complex number:

    [tex] |z| = \sqrt{[Re(z)]^2 + [Im(z)]^2 [/tex]

    So for the specific case of [itex] z = k \in \mathbb{R} [/itex] meaning Im[z] = 0, we get:

    [tex] |z| = |k| = \sqrt{[Re(z)]^2} = \sqrt{k^2} [/tex]

    And that settles it. Okay, I'm feeling pretty stupid now. Do you guys agree I've just answered my own question?
  4. Apr 12, 2006 #3


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    It has to be the absolute value yes, since we defined the (real) function of the square root (k was real here, right?) as a positive number. Take k = -4, then sqrt((-4)²) = sqrt(16) = 4 and not -4, so it was |-4| = 4.
  5. Apr 12, 2006 #4


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    I think this discussion is a very confusing way of looking at this question. Cepheid's professor has taken an approach which is very troublesome. Specifically, there are two square roots to any number. When we write the solution of the quadratic equation with +-, we are emphasizing the fact that we must use both square roots. k2 has two square roots, k and -k, whether or not k is positive or negative. In case k is complex, |k| would be wrong!
  6. Apr 13, 2006 #5


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    "Specifically, there are two square roots to any number."

    No, there aren't. Notice that the professor wrote |k| because k (and so k2)was a real number. That distinction is important. If one were dealing with a differential equation with complex coefficients, then one could argue that we have to deal with "multi-valued" square roots as well as other functions. But this was clearly a differential equation with real coefficients and so the professor was carefully distinguishing between the real and imaginary parts of the complex number. That is, except for the "i" itself, he was working with real numbers. And any positive real number has, by definition, only one square root.
  7. Apr 13, 2006 #6


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    Here is where I strongly disagree. You can define it that way if you want, but it doesn't change the fact that (-2)2=(2)2=4, so that the square root of 4 has two values 2 or -2.
  8. Apr 13, 2006 #7
    The square root function has exactly one value for any given non-negative real number. By definition, that value is the non-negative square root. I know, this is being a bit anal.

    For the simple example in the original post, the answer doesn't depend on whether you put the absolute value in. However, in more complicated examples, it makes a difference, as the OP figured out. So it pays to be anal, sometimes.
  9. Apr 15, 2006 #8


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    If you were asked for the solutions to x2= 3, would you write
    [itex]x= \pm\sqrt{3}[/itex] or just [itex]x= \sqrt{3}[/itex]?

    If you really believed that every number has two square roots, then you would have to say that the latter is the only correct answer since [itex]\sqrt{3}[/itex] indicates both.
  10. Apr 16, 2006 #9
    I thought that every positive real number has two square roots and that the square root sign indicates the positive square root.
  11. Apr 17, 2006 #10
    To get the negative root, it is simply a matter of multiplying the positive root by -1, but only the positive root is THE square root. For the square root relation to be considered a function, it must be one-to-one (pass the vertical line test, loosely speaking). By convention, we take the positive root rather than the negative one to define the square root function.
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