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How Many Solutions To ##z^{1/4}##

  1. Nov 12, 2014 #1
    1. The problem statement, all variables and given/known data
    Is ##z^{1/4}## multi-valued or single-valued? How many possible values can it have in general?

    2. Relevant equations

    ##z## raised to some complex power ##c## is defined as ##z^c = e^{c \log z}##.

    3. The attempt at a solution

    ##
    z^{\frac{1}{4}} = e^{\frac{1}{4} \log z} \iff
    ##

    ##
    z^{\frac{1}{4}} = e^{\frac{1}{4} (\ln|z| + i \arg z)} \iff
    ##

    ##
    z^{\frac{1}{4}} = e^{\ln |z|^{1/4} + \frac{i}{4} (\theta + 2 n \pi)} \iff
    ##

    ##
    z^{\frac{1}{4}} = \sqrt[4]{|z|} e^{i(\frac{\theta}{4} + \frac{\pi}{2} n)}
    ##

    I would say, "yes, it is multi-valued, as each value of ##n## gives you a distinct solution." However, in general, I am unsure of how many solutions there are. For a given ##z##, doesn't ##\sqrt[4]{|z|}## give us four distinct roots; and if we let ##n## be 0, 1, 2, 3, would we get four distinct solutions (n=4 and beyond would give us repetitive solutions)? Furthermore, how do I know the solutions I get from ##\sqrt[4]{|z|}## correspond to the ones gotten by enumerating values of ##n##?
     
  2. jcsd
  3. Nov 12, 2014 #2

    pasmith

    User Avatar
    Homework Helper

    Since [itex]|z|[/itex] is a non-negative real number, both [itex]|z|^{1/4}[/itex] and [itex]\sqrt[4]{|z|}[/itex] always mean the unique non-negative real number [itex]c[/itex] such that [itex]c^4 = |z|[/itex]. Thus there are in general four distinct values of [itex]z^{1/4}[/itex], corresponding to the four distinct values of [itex]n \mod 4[/itex].
     
  4. Nov 12, 2014 #3
    Oh, I see.

    I actually have another one which is also giving me trouble:

    ## (1- \sqrt{3})^{1/5} = e^{ \frac{1}{5} (\ln |1 - \sqrt{3}| + i (\pi + 2 \pi n))} ##

    ## (1- \sqrt{3})^{1/5} = e^{ \ln (1 -\sqrt{3})^{1/5} + i ( \frac{\pi}{5} + \frac{2}{5} \pi n))} \iff ##

    ## (1- \sqrt{3})^{1/5} = (1 - \sqrt{3})^{1/5} e^{i ( \frac{\pi}{5} + \frac{2}{5} \pi n))} \iff ##

    If I let ##n## go through the integers from 0 to 4, I will get the angles ##\frac{\pi}{5}##, ##\frac{3 \pi}{5}##, ##\frac{5 \pi }{5}##, ##\frac{7 \pi}{5}##, and ##\frac{9 \pi}{5}##.

    What would the principal argument be? I see three...I thought the principal argument was that unique angle in the range ##(-\pi,\pi]##, yet I see three.

    EDIT: I don't know why I can't get the latex to render properly. I ran the code in TexStudio and it rendered perfectly. Could anyone spot the issue?

     
    Last edited by a moderator: Nov 12, 2014
  5. Nov 12, 2014 #4

    Mark44

    Staff: Mentor

    You had an extra right brace in there, now removed.
     
  6. Nov 12, 2014 #5

    pasmith

    User Avatar
    Homework Helper



    I see immediately the problem that [itex]1 < \sqrt{3}[/itex], so [itex]|1 - \sqrt{3}| = \sqrt{3} - 1[/itex]. Thus [tex](1 - \sqrt{3})^{1/5} = (\sqrt{3} - 1)^{1/5}e^{i\pi(1 + 2n)/5}.[/tex] There are five distinct roots. Taking [itex]n = 0,1,2[/itex] yields principal arguments of [itex]\frac \pi 5[/itex], [itex]\frac {3\pi}5[/itex] and [itex]\pi[/itex] respectively. This last is the negative real number [itex]-(\sqrt{3} - 1)^{1/5}[/itex]. Taking [itex]n = 4, 5[/itex] gives values outside the range [itex](-\pi,\pi][/itex] so they must be reduced mod [itex]2\pi[/itex] to obtain [itex]-\frac{3\pi}5[/itex] and [itex]-\frac{\pi}5[/itex] respectively.
     
  7. Nov 12, 2014 #6
    Oh, I see. Each distinct has their own principle value. For some reason, I was thinking that only one of the five distinct roots would be the principal argument.
     
  8. Nov 12, 2014 #7
    Okay, I see what the problem is. I was confusing the terms "principal argument" and "principal value."
     
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