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Homework Help: Complex Numbers tips

  1. Feb 20, 2007 #1
    I was wondering if someone can check my solutions and perhaps give me a faster more logical way of working through this question. Thanks.

    1. The problem statement, all variables and given/known data

    If z = 1 + i*root3

    i) Find the modulus and argument of z
    ii) Express z^5 in Cartesian form a + ib where a and b are real
    iii) Find z*zbar
    iv) hence, find zbar to the power of negative 5

    If w = root2 cis (π/4)

    i) Find w/z in polar form.

    ii) Express z and w in cartesian form and hence find w/z in cartesian form.

    iii) Use answers from i) and ii) to deduce exact value for cos(π/12)


    3. The attempt at a solution

    i) Modulus is 2, Argument is π/3

    ii) Using De Moivre's Theorem, 2^5cis(5*(π/3))
    = 32cis(-pie/3)
    Then changing to Cartesian form - 32 (cos(-π/3) + isin(-π/3))
    = 32(0.5 + (root3/2)i)
    = 16 + 16i*root3

    iii) z = 1 + i*root3, zbar = 1 - i*root3
    z*zbar = 4

    iv) Now this part I don't understand the "hence", as in how am I meant to use previous results to get this answer?

    I try - zbar = 2cis(-π/3) -> (Is it true that the "bar" of any complex number is just the same modulus and negative angle?)

    Then zbar^-5 = Using DM Theorem, (1/32)cis(5π/3)
    = (1/32)cis(-π/3)
    Then convert to cartesian - (1/32)(cos(-π/3) + isin(-π/3))
    = (1/32)(0.5 + iroot3/2)
    = 1/64 + iroot3/64


    If w = root2 cis (π/4)

    i) Find w/z in polar form.
    I found it in cartesian - (1+i)/(1 + root3 i) then realising, gives (1 + root3)/4 + (1-root3)/4 * i

    But how do I change to Polar form? I am also stuck on how to get cos(pie/12) as exact value. Thanks.
     
  2. jcsd
  3. Feb 20, 2007 #2

    dextercioby

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    [itex] \sin\frac{-\pi}{3}=-\sin\frac{\pi}{3} [/itex]

    [itex] w=|w|e^{i\theta} [/itex]
     
  4. Feb 20, 2007 #3
    Hello I'm still very basic in this, and I dont understand your notation in the second part. Would you be able to explain further? Thanks!
     
  5. Feb 20, 2007 #4

    dextercioby

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    I thought the complex exponential was the "polar form".
     
  6. Feb 20, 2007 #5

    Dick

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    exp(i*theta)=cis(theta). He may not know the exponential notation.
     
  7. Feb 20, 2007 #6

    Dick

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    "(Is it true that the "bar" of any complex number is just the same modulus and negative angle?)"

    Sure. bar(cis(a))=cis(-a). It's also handy to convince yourself that cis(a)*cis(b)=cis(a+b) and cis(a)/cis(b)=cis(a-b). This is easy to see in the exponential notation. Now note pi/12=pi/3-pi/4. Does that help?
     
  8. Feb 20, 2007 #7
    Yes I have never heard of it, does it make doing the question easier? If so, can you explain?
     
  9. Feb 20, 2007 #8

    HallsofIvy

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    The "hence" part is due to this:
    [tex]\left(z\bar{z}\right)^5= z^5 \bar{z}^5[/itex]
    You know that the left side is 45= 1024 and you have already calculated [itex]z^5[/itex] so you can easily solve for [itex]\bar{z}^5[/itex].
     
  10. Feb 20, 2007 #9

    Gib Z

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    Eulers Formula:

    [tex]e^{ix} = \cos x + i \sin x[/tex]. This lets us write complexs numbers in better ways. O and, De Moirves Theorem is an easy proof with this baby.
     
  11. Feb 20, 2007 #10

    Gib Z

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    O and just incase you want to see how its proved, expand and simplify the taylor series of e^(ix) and see wat you get, or

    let f(x)=cis(x)/e^(ix), find f'(x), once u prove its zero, just find one value and its obvious.

    EDIT: O and iii is much easier with it i think, but i have to go, someone else will tell you.
     
    Last edited: Feb 20, 2007
  12. Feb 20, 2007 #11

    Dick

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    The nice thing about the exponential notation is that it turns an equation like [tex]cis(a)*cis(b)=cis(a+b)[/tex] into [tex]e^{i a} e^{i b}=e^{i (a+b)}[/tex]. Written this way you don't have to remember it as a special rule for the cis function. It's a general property of exponentials. Because cis IS an exponential.
     
  13. Feb 20, 2007 #12

    matt grime

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    Erm, Halls, you're being obtuse for no good reason:

    (z*)^5 = (z^5)*

    * is the same as the over bar.
     
  14. Feb 20, 2007 #13

    HallsofIvy

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    Well, that happens occaissionally! Perhaps more often that I would like!
     
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