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How to represent this complex number?

  1. Mar 13, 2016 #1
    1. The problem statement, all variables and given/known data
    Finding "polar" and "rectangular" representation of a complex number?

    Make a table with three columns. Each row will contain three representations of a
    complex number z: the “rectangular” expression z = a + bi (with a and b real); the “polar”
    expression |z|, Arg(z); and a little picture of the complex plane with the complex number
    marked on it. There are five rows, containing, in one column or another, the following
    complex numbers:


    (iv) A sixth root of 1 with argument θ such that 0 < θ < π/2

    2. Relevant equations
    I don't know if de moivre's theorem for Nth roots of complex numbers is relevant here

    3. The attempt at a solution
    I know that the modulus of Z = r = 1^(1/6) = 1 but I don't know how to find the Argument of Z

    Z = 1^(1/6) e^(iθ)
     
    Last edited by a moderator: Mar 13, 2016
  2. jcsd
  3. Mar 13, 2016 #2

    LCKurtz

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    Hint: ##(re^{i\theta})^6 = r^6e^{i6\theta}##
     
  4. Mar 13, 2016 #3

    Merlin3189

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    I don't know anything about this, but now someone has answered, I'll tell you my way.
    Start with the third column! Sketch the diagram and use trig to work out the other two columns.
     
  5. Mar 13, 2016 #4

    SammyS

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    If ##\ z=\sqrt[6]1 \,,\ ## then ##\ z^6 = 1\ .\ ## Right?

    What is the argument of 1 ?
     
  6. Mar 13, 2016 #5
    wouldn't the argument of 1 just be zero since it doesn't have an imaginary part?
     
  7. Mar 13, 2016 #6

    Merlin3189

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    That's one answer. What about the others?
     
  8. Mar 13, 2016 #7
    Isn't that the
    Isn't that the only choice since 0 < θ < pi/2?
     
  9. Mar 13, 2016 #8

    Merlin3189

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    Since you say ##\ z^6 = 1\ .\ ## presumably ##6\times Arg(z)\ =\ Arg(1)## or ##6\times θ\ =\ Arg(1)##
    That does not mean Arg(1) must be between 0 and π/2

    if you take Arg(1) as 0, then θ = 0/6 = 0. But you are asked for θ >0

    Edit: If 0 < θ < π/2 then ##6\times θ\ =\ Arg(1)## might mean 0 < Arg(1) < 6π/2
     
    Last edited: Mar 13, 2016
  10. Mar 13, 2016 #9
    I already know the answer, which is

    1^(1/6) = 1*e^(i*pi/3) = 1/2 +i*√3/2. So Arg(z) = pi/3

    but I just can't see how to get there :/
     
  11. Mar 13, 2016 #10
    Mind if I hop in? You have that [itex] Z = 1^{1/6} [/itex]. 1 is a complex number so it can be represented in the form [itex] Z = (re^{i\theta})^{1/6} [/itex]. As you said 1 has no imaginary part so r = 1. Now you have [itex] Z = e^{\frac{i\theta}{6}} [/itex]. Notice that the argument is effectively reduced by a factor 6,So there are other values for the argument within your bounds. You had 0 as one solution for theta, in trigonometry, how would you shift the argument of a function and still retain the same answer? Hope this helps.
     
  12. Mar 13, 2016 #11
    Are you referring to the fact that you can add 2pi to the argument and have the same answer?
     
  13. Mar 13, 2016 #12
    Yes
     
  14. Mar 13, 2016 #13

    LCKurtz

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    You could also ask yourself what angle could I add up 6 times to get ##2\pi##?
     
  15. Mar 13, 2016 #14
    So if r = 1 then Z = e^(iθ/6) and 0 < θ/6 <pi/2, and since Z= 1^1/6 the complex number has no imaginary part. Therefore the argument of Z has to be either 0/6 or 2pi/6?

    but since we have the restriction 0 < θ/6 < pi/2 , it must be 2pi/6 = pi/3?
     
  16. Mar 13, 2016 #15
    That looks good to me.
     
  17. Mar 13, 2016 #16
    One more question though, since Z = 1^(1/6), we said that the complex number had no imaginary part. This is true for argument of Z at θ = 0, but not at θ =pi/3. So how can we maintain equivalent answers by adding 2pi when doing so gives us an imaginary number we didn't have to begin with?
     
  18. Mar 13, 2016 #17
    This is how I understand it: while it is true that 1 has no imaginary part, it is still a complex number in the complex plane. It is equivalent to saying 1+ 0i. A complex number represented as a + bi. From this cartesian form you have that [itex] r = \sqrt{a^2 + b^2} [/itex]. Both our answers, when represented in cartesian form, will give the same value of r. This diagram from Wolfram Alpha may help.( Both answers give the same value of r, which is the radius of the circle in the complex plane). Hope this helps, perhaps someone else can explain it better than I have :smile:
     

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  19. Mar 13, 2016 #18
    Ok I think I understand it a little better now. Thank you all for your help!
     
  20. Mar 14, 2016 #19

    Merlin3189

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    Agree with bigguccisosa.

    Even ##\sqrt{1}## has the same issue. ##z^2\ =\ 1## so ##2\times Arg(z)\ =\ Arg(1)## and, at first, ##Arg(z)\ =\ 0##
    Only when you let Arg(1) be 2π, do you get 2 x Arg(z) = 2π and Arg(z) = π giving you the second (real) sqrt of 1, which is -1

    To get all 6 roots for ##\sqrt [6] {1}## you need to consider 6 values for Arg(1) suc as -6π to +6π (that's 7 actually: miss off either 6π or -6π, but not both) (or use 0 to 10π, but for me something involving 6 is more intuitive!)

    But for me the easiest way is just to use the diagram that bigguccisosa showed to get the arguments of the roots. Here the unit circle divided into 6 is 60o or π/3 just by inspection (and giving roots at n x 60o. or nπ/3 )
    And the a + ib coordinates are equally easy (for 3rd, 4th, 6th and 8th roots, because the tangents are well known fractions) as ## \pm \frac {1}{2} \ \pm \frac {\sqrt(3)}{2}i \ \ \ and \pm 1 + 0i##
     
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