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Homework Help: Complex roots

  1. Sep 21, 2010 #1
    1. The problem statement, all variables and given/known data

    Z^5 = -1

    What are the five roots?

    2. Relevant equations

    3. The attempt at a solution

    I do not know how to start this problem. I am given a hint to put -1 into the form Ae^(i*delta), but I am unsure as to how I can do that. Any help would be much appreciated.
  2. jcsd
  3. Sep 21, 2010 #2


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    e^{ix}=\cos x+i\sin x
  4. Sep 21, 2010 #3
    Ya, I know. That was the hint I was given. However, I don't know exactly how to apply it in this situation. How do I get five roots from that equation?
  5. Sep 21, 2010 #4


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    -1=\cos x+i\sin x\Rightarrow \cos x =-1 \quad \sin x =0.
    What could x be?
  6. Sep 21, 2010 #5
    he could also use

    z = x + iy

    z^5 = x^5 ... iy^5
    (fill in the rest)

    to remove the theta
  7. Sep 21, 2010 #6
    X could be any odd integer multiple of pi (i.e. pi, 3pi, 5pi, 7pi....). Is that what you are talking about?
  8. Sep 21, 2010 #7


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    Almost, not any odd integer of pi, but a specific one.
  9. Sep 21, 2010 #8
    I'm sorry. I don't see why it has to be a specific one. All of those work so how do I know which particular one is correct?
  10. Sep 21, 2010 #9
    There is another way too.
    [tex] r^{5}e^{5 thetha } = e^{i \pi + 2n \pi}[/tex]

    This way you can get all the values of for the angle from a " formula".
  11. Sep 21, 2010 #10

    -1 + i*0 = cos(x) + i*sin(x)


    cos(x) = -1

    sin(x) = 0

    To satisfy both these conditions, X can be any odd integer multiple of pi.

    cos(n*pi) = -1 for n= 1, 3, 5, 7, ...

    sin(n*pi) = 0 for n= 1, 3, 5, 7, ...

    I don't see how there can be a single specific answer.

    I also don't understand where I am going to get five roots from.
  12. Sep 21, 2010 #11
    You are correct.

    [tex] cosx = -1 \Rightarrow x = \pi +2n\pi[/tex]

    [tex]x = \pi +2n\pi [/tex]

    Now that you have
    [tex]-1 = e^{i \left(\pi + 2n \pi \right)}[/tex]

    You also have [tex] r^{5}e^{i5 \theta } = e^{i \left(\pi + 2n \pi \right)}[/tex]

    What is r and theta ?
    [tex] r = ?[/tex]
    [tex] \theta = ? [/tex]
  13. Sep 21, 2010 #12


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    What the other posters have been trying to guide you through is a two-step process. The first step is to find a way to express -1 in the form [itex]re^{i\theta}[/itex] - that is, what values of r and θ will make the expression [itex]re^{i\theta}[/itex] equal to -1? For this part, you use Euler's identity,
    [tex]e^{i\theta} = \cos\theta + i\sin\theta[/tex]
    As you have found, there is not just one solution for θ, but a whole infinite sequence.

    Once you've found the value(s) of r and θ that correspond to -1, the second part of the problem will be to figure out what complex numbers to the fifth power equal -1. You can use the usual property of exponents,
    [tex](a^b)^c = a^{bc}[/tex]
    Since you have an infinite set of possible values for θ for -1, go through them one by one and for each one, find the complex number (in polar form) that, when raised to the fifth power, gives you your r times e for that θ. Obviously you cannot do the entire infinite sequence, but do enough of them to see the pattern.

    After you're done with that, if you convert the complex numbers you found from polar form back to Cartesian form (x+ iy), you should find that there are only five different values. Those are the roots you're looking for.
  14. Sep 21, 2010 #13
    Okay, thanks for the explanation diazona. However, let me show you my thought process.

    So I have already found that when theta=pi, 3pi, 5pi, 7pi, ... ; e^(i*theta) will equal -1.

    I also think that the only possible value for r is 1.

    So then you say I need to find what complex numbers when raised to the fifth power give me -1.

    So, (e^(i*theta))^5 = e^(i*5*theta).

    My question is regarding the previous step. If I am to plug in all the possible values of theta (pi, 3pi, 5pi, 7pi, ...) then when I multiply by five I will still get an odd multiple of pi and it will always result in -1.
  15. Sep 21, 2010 #14
    Either you don't need my help or you don't want to listen( or read) to me.

    I would take the former as the case.
  16. Sep 21, 2010 #15
    Actually it's neither of those. Since I don't know how to do this, I can't just look at an equation and know how to navigate my way to the final answer.

    Yes, I see your equation: (r^5)(e^(i*5*theta))=-1

    This works when r=1 and theta=pi so this is one solution.

    This also works when r=-1 and theta=0 so this is a solution.

    What I was getting at was do I need 5 unique r values and 5 unique theta values because it also works when r=-1 and theta=2pi.
  17. Sep 21, 2010 #16
    Good. You get the general idea.

    Look at this
    [tex] r^{5}e^{i5 \theta } = e^{i \left(\pi + 2n \pi \right)}[/tex]

    So far you have deduced that
    [tex] r= 1[/tex]

    We see that
    [tex] 1^{5}e^{i5 \theta } = e^{i \left(\pi + 2n \pi \right)}[/tex]
    [tex]e^{i5 \theta } = e^{i \left(\pi + 2n \pi \right)}[/tex]

    This means

    [tex] 5 \theta = \pi + 2n \pi [/tex]

    Can you solve for theta and then try n = 0 ,1, 2,3,4 ?

    What values do you get for theta?
  18. Sep 21, 2010 #17
    Actually, I forgot to point out that r is defined to be positive.

    So you can't pick r = -1.
  19. Sep 21, 2010 #18
    Solving for theta and plugging in n=0,1,2,3...

    I get:

    theta = (1/5)*pi
    theta = (3/5)*pi
    theta = (5/5)*pi
    theta = (7/5)*pi
  20. Sep 21, 2010 #19

    You are missing n=4.

    Anyway, what you have here are the arguments of complex numbers which are all roots of -1 .

    These are your answers.

    EDIT :
    To see this

    [tex] \left(e^{\frac{i\pi}{5}}\right)^{5} = e^{i\pi} =-1 [/tex]

    And so on...
  21. Sep 21, 2010 #20
    Okay, I see it now. Ya, I forgot n=4. I see what you mean about only having 5 answers because when n=5 it just cycles back around to (1/5)*pi.

    Thank you so much for your patience. I'm sure this was very frustrating for you.
  22. Sep 21, 2010 #21
    Yes, it's just as you say; it cycles back.

    Not at all.

    I myself had a hard time understanding why a single "number" would have 5 roots when I was introduced to the idea.
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