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Square root of i

  1. Apr 29, 2005 #1
    Does i, the imaginary number, have a square root? This was bothering me for a while, then I thought I happened upon a simple solution, but have since forgetten.

    [tex]\sqrt{i}=?[/tex]
     
  2. jcsd
  3. Apr 29, 2005 #2

    Hurkyl

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    It does if you can solve the equation z2 = i.
     
  4. Apr 29, 2005 #3
    I'm pretty sure the answer is of the form a+bi where a,b are non-zero.
     
  5. Apr 29, 2005 #4

    James R

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    The square roots of i are:

    [tex]z = \frac{1}{\sqrt{2}}(1 + i), -\frac{1}{\sqrt{2}}(1 + i)[/tex]

    You can check by squaring these numbers.
     
  6. Apr 29, 2005 #5
    [tex]z= \pm\frac{1}{\sqrt{2}}\left(1+i)[/tex]

    has the property [itex]z^2=i[/itex]. You can find this by solving

    [tex](a+bi)^2=i[/tex]

    for [itex]a[/itex] and [itex]b[/itex] real.
     
  7. Apr 29, 2005 #6
    Excellent, I got to finding that (1+i)^2=2i. So that makes sense.
     
  8. Apr 29, 2005 #7
    and by the way, you shouldn't need to ask "does i have a square root." The complex numbers are algebraically closed, ie. every nonzero complex number has exactly [itex]n[/itex] [itex]n^\mbox{th}[/itex] roots.
     
  9. Apr 29, 2005 #8
    Interesting. "Algebraically closed"? Does that just mean the set is closed under the operation of taking roots?
     
  10. Apr 29, 2005 #9

    Hurkyl

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    Even better -- every nonconstant polynomial has a root!
     
  11. Apr 29, 2005 #10
    De Moivre's Theorem...


    There are two solutions for the square root of i using De Moivre's Theorem:

    [tex] w_k = \text{cis} \; \frac{90^{\circ} + 360^{\circ} k}{2} [/tex]
    [tex] w_0 = \text{cis} \; 45^{\circ} = (\cos 45^{\circ} + i \sin 45^{\circ}) [/tex]
    [tex] (\cos 45^{\circ} + i \sin 45^{\circ}) = \left( \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i \right) [/tex]

    [tex] w_1 = \text{cis} \; 225^{\circ} = ( \cos 225^{\circ} + i \sin 225^{\circ}) [/tex]
    [tex] ( \cos 225^{\circ} + i \sin 225^{\circ}) = \left( - \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} i \right) [/tex]
    [tex] \boxed{w_0 = \frac{\sqrt{2}}{2} (1 + i)} [/tex]
    [tex] \boxed{w_1 = - \frac{\sqrt{2}}{2} (1 + i)} [/tex]
    [tex] \boxed{z = \pm \frac{\sqrt{2}}{2} (1 + i)} [/tex]

    Note that this solution is irrational with a root in the denominator:
    [tex]z = \pm\frac{1}{\sqrt{2}}\left(1+i)[/tex]

    Any root of i can be solved using De Moivre's Theorem.

     
    Last edited: Apr 29, 2005
  12. Apr 29, 2005 #11
    And to see why Hurkyl's comment is significant, remember that over the reals, for example, not every polynomial has a root - for example, [itex]x^2+1[/itex] has no real roots, and it can't be factored over the reals. Over the complex numbers, every polynomial has a root, and this amounts to saying that every monic complex polynomial can be factored to the form

    [tex](x - \alpha_1)^{e_1} \ \cdots \ (x - \alpha_n)^{e_n}[/tex]

    for some positive integers [itex]e_i[/itex] and distinct complex numbers [itex]\alpha_i[/itex], in a unique way (up to ordering of the factors).

    The fact that every nonzero complex number has [itex]n[/itex] nth roots is a little easier, though, as long as you know that every nonzero complex number [itex]z[/itex] can be represented in a unique way in the form [itex]re^{i\theta}[/itex] for some [itex]\theta \in \left[0, 2\pi), \ \mathbb{R} \ni r > 0[/itex], and that [itex]e^{i\theta}[/itex] is 2[itex]\pi[/itex]-periodic. Then you can write

    [tex]z = re^{i\theta} \Longrightarrow \left(\sqrt[n]{r}e^{i\left(\frac{\theta + 2k\pi}{n}\right)}\right)^n = z, \ \mathbb{Z} \ni k \in \left[0, n-1\right][/tex]

    ie. z has at least [itex]n[/itex] nth roots. It remains to be proved that there are exactly n of them, but you can work on that on your own~
     
    Last edited: Apr 29, 2005
  13. Apr 30, 2005 #12
    From Orion1:

    [tex] w_1 = \text{cis} \; 225^{\circ} = ( \cos 225^{\circ} + i \sin 225^{\circ}) [/tex]


    Is the cis() function equivalent to cos() + i sin()

    ?
     
  14. Apr 30, 2005 #13

    jcsd

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    Yes it is, but no-one bothers to use it any more as cis x = e^ix which is just as easy to write.
     
  15. Apr 30, 2005 #14
    Orion1: Any root of i can be solved using De Moivre's Theorem.

    I just want to add that, if its the inverse of a power of 2, it can be solved in terms of repeated square roots.
    Looking at the 4th root of i, we have, in degrees, that the

    sin(22.5) = [tex]\sqrt{\frac{1-cos(45)}{2}}[/tex]

    This allows us to find the sin(22.5) = [tex]\sqrt{\frac{2-\sqrt{2}}{4}}[/tex]

    So if we apply DeMovre's theorem, we can find the answer in term of surds.

    This method. of course, can be applied to the next level to give

    sin(11.25) =[tex](1/2)\sqrt{2-\sqrt{2+\sqrt{2}}}[/tex]

    It can be added that these repeated square root equations are, as Gauss had pointed out, the only constructable ones using a ruler and compass according to the method of the Greeks.
     
    Last edited: May 1, 2005
  16. May 1, 2005 #15
    i dont know if my answer is of any relevance since i am always ignored or insulted.
    but i did a work of mine and got +or-[i/rt2] *[1-i].you can tell me if i am wrong
     
  17. May 1, 2005 #16

    HallsofIvy

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    I don't know that anyone is insulting you (and I certainly can't tell if anyone is ignoring you but your post on an different "quadratic formula" got a number of responses!) but you are getting quite a bit of criticism about careless expression. Here you say "i did a work of mine and got +or-[i/rt2] *[1-i].you can tell me if i am wrong". No one can possibly say if [tex]\pm\frac{i}{\sqrt{2}}*(1-i)[/tex] is right or wrong because you haven't told us what you were working on or what it is supposed to be the answer to! If you are suggesting it as an answer to "what are the square roots of i?" which was the original question, couldn't you just square it yourself and see?
     
  18. May 1, 2005 #17
    keep it simple-

    i^(1/2) = (-1)^(1/4)
     
  19. May 1, 2005 #18

    cepheid

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    I don't see how that helps particularly, the RHS is not any easier to do than the LHS. But if you know how to do the former, you also know how to do the latter anyway! Besides, there are four fourth roots of one and only two square roots of i (see post #7). Only the two common to each will satisfy that equation. Just do this, always:

    [tex] z = re^{i\theta} [/tex]

    [tex] z^{\frac{1}{m}} = \sqrt[m]{r} e^{i(\theta + 2k\pi)/m [/tex]

    [tex] k = 0, 1, ... , m-1 [/tex]

    "r" being a modulus of course, [itex] \sqrt[m]{r} [/itex] refers to the positive, real root.
     
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