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5 degree equation contains imaginary value?

  1. Jun 20, 2017 #1
    1. The problem statement, all variables and given/known data

    15_Mat_B_-_1.5.png
    2. Relevant equations
    polynomial equation $$(a+b)^2 = \sum_{k=0}^{n} \binom nk a^{n-k} b^k $$

    3. The attempt at a solution
    i get $$ \frac {1} {32} + \frac {5} {16}\frac {3^{0.5}i} {2} +\frac {10} {8}\frac {3i^2} {4} + \frac {10} {4}\frac {3^{1.5}i^3} {8} + \frac {5} {2} \frac {9 i^4} {16} + \frac {9.3^{0.5}i^5} {32} $$

    and it didn't work out. can anyone show me other way to solve it?
     
  2. jcsd
  3. Jun 20, 2017 #2
    Are you allowed to use polar form ?
     
  4. Jun 20, 2017 #3

    Mark44

    Staff: Mentor

    I didn't verify your work above, but you should replace i^2 by -1, and i^3 by -i, and so on.

    This problem is much easier if you use @Buffu's suggestion, assuming you know about polar form for complex numbers.

    BTW, your "relevant equation" isn't relevant, as it's wrong. With an exponent of 2 on the left side, the summation should be ##\sum_{k=0}^2 \binom 2 k a^{2-k} b^k##.
     
  5. Jun 20, 2017 #4
    why should I replace that? i^2 = -1 ? i^3 = -i? why?

    yes, what exactly the purpose of using polar form?
    w = 1/2 + ( (root3) / 2) i
    r = 1, tetha = 60degrees??
     
  6. Jun 20, 2017 #5

    Mark44

    Staff: Mentor

    Because by doing this you can reduce the 6 terms you showed to only 2 terms.
    Multiplication of complex numbers is much easier in polar form. If ##z_1 = r_1e^{i\theta_1}## and ##z_2 = r_2e^{i\theta_2}##, then ##z_1z_2= r_1r_2e^{i(\theta_1 + \theta_2)}##

    By the way, this Greek letter -- ##\theta## -- is theta, not tetha.
     
  7. Jun 20, 2017 #6
    i see..
    replacing that not changing the value, so i^2 equal to -1 ?
    i^3 equal to -i? what theory is used?
    so i^4 = (i^2)^2 = (-1)^2 = 1?
    i^5 = i^4 . i = i
     
  8. Jun 20, 2017 #7

    Mark44

    Staff: Mentor

    Yes to all. ##i^2 = -1## by definition. ##i^3 = i^2 i = -1 i = -i##, and so on.
     
  9. Jun 20, 2017 #8
    I got the answer, by using conventional method. I would like to know how polar form can solve this problem... @Buffu
    @Mark44 do you know how too?
     
  10. Jun 20, 2017 #9

    Mark44

    Staff: Mentor

    Leaving the complex number in Cartesian (or rectangular) form will work, but it takes more work.

    In polar form ##\omega = \frac 1 2 + \frac {\sqrt 3} 2 i = e^{i \pi/3}##. Note that because ##|\omega| = 1## the r factor in the polar form is just 1, and I don't show it.
    ##\omega^5 = 1^5 e^{i5 \pi/3}##, which represents a complex number whose angle is ##5\pi/3## and whose magnitude is (still) 1.This exponential polar form can be converted into trig polar form, because ##e^{i\theta} = \cos(\theta) + i\sin(\theta)##, due to Euler's Formula.

    So all you have to do for this problem is to evaluate ##\cos(5\pi/3)## and ##\sin(5\pi/3)##.
     
  11. Jun 20, 2017 #10
    z = re = r(cosθ + i*sinθ). In this case r = 1 and θ = π/3
    z5 = r5ei5θ = ei5π/3 = e-iπ/3 = 1/2 - i√3/2
     
  12. Jun 20, 2017 #11
    I think you should have used ##\omega^3 = -1##, then you only need to evaluate ##\omega^2## which is easier .
     
    Last edited: Jun 20, 2017
  13. Jun 20, 2017 #12
    How w = 1? Isnt w^2 = -1 by definition?
     
  14. Jun 20, 2017 #13
  15. Jun 20, 2017 #14
  16. Jun 20, 2017 #15
    No I mean ##\omega^3 = -1 \implies (-\omega)^3 = 1##. Look at the link in my previous you can see three cube roots of ##1##.
     
  17. Jun 20, 2017 #16
    Hmm on the link i saw the property is w^3 = 1
    W^5 = w^2
    W^5 = 1 + root3 i + 3/4 i^2
    While i^2 = -1
    = 1 + root3 i - 3/4
    = -1/4 + 4root 3 i / 4

    It didnt work out.
    If w^3 = -1
    Then w^5 = 1/4 - 4root 3 i/4
     
  18. Jun 20, 2017 #17
    How can ei5π/3 = e-iπ/3 ? @mjc123
     
  19. Jun 20, 2017 #18
    In that page ##\omega^\prime = \dfrac{- 1 \pm \sqrt{3} i}{2}, 1## whereas here ##\omega =\dfrac{+1 + \sqrt{3} i}{2} = -\dfrac{1 + \sqrt{3} i}{2} = -\omega^\prime##.

    ##\exp(5i\pi/3) = \cos 5\pi/3 + i \sin 5\pi/3 = 1/2 + -\sqrt{3}i/2 = \exp(-\pi i/3)##
     
  20. Jun 21, 2017 #19
    Adding a multiple of 2π to the angle brings you back to the same point. So e = ei(θ+2nπ)
    So ei5π/3 = ei(5π/3 -2π) = e-iπ/3. Draw the Argand diagram to see this.
     
  21. Jun 21, 2017 #20
    why do I have to do it? If it's still theta?
     
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