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What is the square root of i^2

  1. Oct 4, 2006 #1
    what does [tex]\sqrt{i^2}[/tex] equal to? is [tex]\sqrt{i^2} = i[/tex] or [tex]\sqrt{i^2} = \pm i[/tex]?
     
    Last edited: Oct 4, 2006
  2. jcsd
  3. Oct 4, 2006 #2
    im guessing its plus&minus
     
  4. Oct 4, 2006 #3

    matt grime

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    The square root is usually taken to be the principal branch of the function. I.e. the root whose argument lies in [0,pi), so you would conventionally obtain i.

    In general the n'th root of a complex number is taken to be the principal root unless otherwise stated: the root with argument in the interval [0,2pi/n)
     
  5. Oct 4, 2006 #4
    I gave you the usual definition of the square root function in your other thread. You can use that to figure out what it is.
     
  6. Oct 5, 2006 #5
    you said, [itex]\sqrt{z} = z^{1/2} = \sqrt{r}e^{i\theta/2}.[/itex]

    but if [tex]z = x + iy[/tex], then [tex]r = \sqrt{x^2 + y^2}[/tex] and [tex]\theta = \tan^{-1}\left( y \over x \right)[/tex]

    if [tex]z = -1 = -1 + 0.i[/tex], then x = -1, y = 0.

    [tex]r = \sqrt{\left(-1\right)^2} = \sqrt{1} = 1[/tex]

    [tex]\theta = \tan^{-1}\left( {0} \over {-1} \right) = 0[/tex]

    [tex]\sqrt{z} = \sqrt{-1} = \sqrt{r}e^{i\frac{\theta}{2}} = \sqrt{1}e^{0} = 1[/tex]

    but where did i go wrong?
     
    Last edited: Oct 5, 2006
  7. Oct 5, 2006 #6

    matt grime

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    Because arctan only returns values in the region -pi/2 to pi/2, so it is wrong to say that theta is arctan(y/x) alone. You must also use your common sense to see that the argument is actually pi, not zero. It is correct to say that tan(theta)=y/x, but that is *not* the same as theta =arctan(y/x) at all.
     
  8. Oct 5, 2006 #7
    yeah, i can see that [tex]\theta = \pi[/tex] from the position of z = -1 in the argand plane. so,

    [tex]\sqrt{z} = \sqrt{-1} = \sqrt{r}e^{i\frac{\theta}{2}} = \sqrt{1}e^{i\frac{\pi}{2}} = 1.(\cos{\frac{\pi}{2}} + i.\sin{\frac{\pi}{2}}) = i[/tex]

    am i right?
     
  9. Oct 7, 2006 #8

    This is my assumption:

    [tex]\sqrt{i^{2}} = \sqrt{-1} = i[/tex]

    [tex]i^{2}[/tex] is equal to -1 and square root of -1 is defined as i.
     
  10. Oct 8, 2006 #9
    That expression for [itex]r[/itex] works for all [itex]z[/itex], but the one for [itex]\theta[/itex] only works for [itex]x>0[/itex], assuming you're using the usual branch of arctan. In general finding the polar representation of a complex number involves solving

    [tex]\sin{\theta} = \frac{y}{r}[/tex]
    and
    [tex]\cos{\theta} = \frac{x}{r}[/tex]

    simultaneously (taking [itex]\arctan (y/x)[/itex] makes the possible solutions obvious though!).
     
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