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Changing complex numbers in form a+bi

  1. Jun 7, 2012 #1
    1. The problem statement, all variables and given/known data
    Here is the problem: ([itex]\sqrt{}6[/itex](cos(3pie/16)+i sin(3pie/16)))^4


    2. Relevant equations

    After 360*=2pier Do you add 2pier to the next degree which would be 30*=pie/6?

    3. The attempt at a solution
     
  2. jcsd
  3. Jun 7, 2012 #2
    In situations such as these, it is usually easiest to convert to the exponential form and solve from there, then convert back to a+bi form.
     
  4. Jun 7, 2012 #3

    vela

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    What do you mean by this? By "2pier" do you mean ##2\pi## radians or ##2\pi r##, the circumference of a circle? What is "the next degree"? After 360 degrees, the next degree is 361 degrees, no?
     
  5. Jun 7, 2012 #4
    The greek letter [itex]\pi[/itex] is just spelled as "pi". It is not the same as the dish.
     
  6. Jun 7, 2012 #5
    Sorry, I couldn't find the 2 "pie" r.

    No worry guys I figured it out. Wish I could post pictures... I would show you how I solved it.

    EDIT: Sweet I can post pics now!
     
  7. Jun 8, 2012 #6
    Here is the problem
    0607021250.jpg
    1-2.jpg
    9.jpg
    10.jpg
     
  8. Jun 8, 2012 #7
    Per daveb, I too think it would be much easier to convert to polar form first and then convert back. The reason is that the exponential lends itself to being raised to a power much more easily than the rectangular form.

    What you get is

    [tex](\sqrt{6} e^{3\pi i/16})^4 = 36 e^{3\pi i/4}[/tex]

    This is what you got, after all, but you can do it in one line as opposed to resorting to a bunch of trig identities to square the rectangular form.
     
  9. Jun 8, 2012 #8
    Apologies for the quality of the pictures...
     
  10. Jun 10, 2012 #9
    Look up de Moivre's Theorem. It'll narrow down the tedious computation with multplying it out.
     
  11. Jun 10, 2012 #10
    $$(cos(x)+i~sin(x))^{n} = cos(nx)+i~sin(nx)$$
     
  12. Feb 6, 2014 #11
    Whoa... I was a member while taking trig, crazy!
     
  13. Feb 7, 2014 #12
    So much easier when using Euler's formula. I can do it in my head rather quickly.
     
  14. Feb 7, 2014 #13
    Good for you.
     
  15. Feb 8, 2014 #14
    Just sayin' - if you recognize this

    c9f2055dadfb49853eff822a453d9ceb.png

    then the problem become easy to do. There are plenty of ways to skin the cat with this problem.
     
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