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+ and - roots

  1. Jul 19, 2015 #1
    1. The problem statement, all variables and given/known data

    If you take a 4th root of something, is that answer also plus or minus, just like if you are taking a square root?

    Ex. ##4^{\frac 1 2} = ±2##

    So following that logic, would that mean that:

    ##16^{\frac 1 4} = ±2## as well?

    Meaning, is the ± still present if you take a 4th root?


    2. Relevant equations


    3. The attempt at a solution
     
  2. jcsd
  3. Jul 19, 2015 #2

    Mark44

    Staff: Mentor

    While 2 and -2 are square roots of 4, the expression ##\sqrt{4}## denotes the principal (i.e., positive) root, or +2. So ##4^{1/2}## would also be +2.
    The principal fourth root of 16 is 2. There are three other fourth roots: -2, 2i, and -2i.
     
  4. Jul 19, 2015 #3
    Okay, so I took the 4th root of -1 using a CAS, and ended up getting the result of

    ## r = \frac {\sqrt(2)} {2} +\frac {\sqrt(2)} {2}i ##

    So is there any other answer besides this, or is this the single and only ?
     
  5. Jul 19, 2015 #4

    Mark44

    Staff: Mentor

    There are four fourth roots, spaced 90° apart (##\pi/2## radians). I don't know if there is a principal fourth root of a negative number, similar to what I said before about the fourth root of 16.

    Usually the question is asked as, "Find all of the fourth roots of <N>."
     
  6. Jul 19, 2015 #5

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    The fourth roots of -1 satisfy the equation ##r^4 = -1## or ##r^4 + 1 = 0##. Since you have a fourth-degree polynomial, there are going to be four solutions.

    To find the square roots of 4, you want to solve the equation ##z^2 = 4##. If you express both sides in polar form, with ##z = re^{i\theta}##, you have
    $$r^2 e^{i2\theta} = 4e^{i0},$$ which has solution ##r=2## and ##\theta=0##, which corresponds to ##z=2##. But you could also write the righthand side as ##4e^{i2\pi}##, which gives you a second solution, ##r=2## and ##\theta=\pi##, which corresponds to ##z=-2##.

    You can use the same method to find all the fourth roots.
     
  7. Jul 19, 2015 #6
    Okay, thank you both, I understand how to do this now and have applied it to my problem, but my problem is differential calculus based, so I will post that in the calculus and above section.
     
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