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Homework Help: Using Argument and DeMoivre's theory

  1. Mar 13, 2010 #1
    Hello everyone,
    I'm having a bit of trouble with using the argument function with DeMoivres formula. I have the question:
    z8= 16
    and am meant to find the solution using DeMoivre's formula (zn=rn(cosn(Θ) +isinn(Θ)) ). The problem is, I have no idea what an argument function is or how to find it. I've read around a bit and found that root(a2+b2)= r and that tan-1(b/a) =arg

    but as you can see, I only have a (16) so that b=0, so is the argument then, tan-1(0)? I have several problems and they all have a but no b. However, in the answer section, they all have arguments. What the heck am I doing wrong?


    2. Relevant equations
    I can do this one:
    z3=-1 as it has no argument (though I don't know why), and is just a matter of plugging in numbers.
    z3=-1
    z3=1(cos(Θ)+isin(Θ)
    z=3√(1)(cos(Θ)+isin(Θ))1/3
    z=1(cos(2kπ)+isin(2kπ)1/3
    z=1(cos(2kπ/3)+isin(2kπ/3)

    and then, since it has no argument, k can equal 0, +/-1, +/-2, +/-3 etc..., you plug in the k and solve.



    3. The attempt at a solution

    I haven't gotten very far with this one

    z8=16
    z8= 168(cos8(Θ)+isin8(Θ)) or if I do it the other way
    z=8√(16)(cos(2kπ)+isin(2kπ))1/8
    z=8√(16)(cos(2kπ/8)+isin(2kπ/8))

    and I can't insert anything for k, because I don't know the argument. I know I sound like a real novice at math (I am), but I hope someone can help me! Thank you so much in advance!
     
  2. jcsd
  3. Mar 13, 2010 #2

    phyzguy

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    Science Advisor

    Imagine plotting a complex number z = x + i y in the complex plane, with the x coordinate the real part and the y coordinate the imaginary part. The argument is then the angle made between the real axis and a line drawn between the origin and the point z. Any positive real number has an argument of zero, and any negative real number has an argument of [tex]\pi[/tex]. A positive imaginary number has an argument of [tex]\pi[/tex]/2. So, don't say that 1 has "no argument", say that it has an argument of zero.
     
  4. Mar 13, 2010 #3
    Hey, thanks so much for your reply!

    Using what you told me, I've managed to get to the next question, but I get stuck here. I'm not sure what I am doing wrong. Here's where I am so far:

    z6=1+i
    z=r1/6(cosΘ+isinΘ)1/6
    z=6√2(cos(Θ/6)+isin(Θ/6))
    z=6√2(cos(π/12)+isin(π/12))

    and so

    zk=6√2 ei(π+2πk/12)

    The answer I need to get is:

    zk=21/12ei(π+8πk/24)

    I don't understand at all what happened. Am I still doing the argument incorrectly? Thanks in advance!
     
  5. Mar 13, 2010 #4

    phyzguy

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    Science Advisor

    I think you're still doing the argument incorrectly. It looks like you have the right answer for the magnitude, since the sixth root of sqrt(2) = 21/12. But what is the argument of 1+i? You should get tan-1(1/1) = tan-1(1) = [tex]\pi[/tex]/4
     
  6. Mar 13, 2010 #5
    I finally got it! Thanks so much!
     
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