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Math Help - De Moivre's theorem problem :(

  1. #1
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    De Moivre's theorem problem :(

    The question states:

    Use the polar form and De Moivre's Theorem to simplify:
    (a) (1+i)^5
    _______

    1 - i

    So I categorised each as z1 and z2, worked out the modulus and argument of each, ( |z1| = root 2, |z2| = root 2, arg z1 = pie/4, arg z2 = - pie/4)

    Carried out de moivre's theorem and got as far as the step

    4cis(5pie/4)
    __________

    cis(-pie/4)

    It would be much appreciated if anyone could help me finish it off or correct me if I've gone wrong somewhere along the line. Cheers
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  2. #2
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    Quote Originally Posted by scott.b89 View Post
    The question states:

    Use the polar form and De Moivre's Theorem to simplify:
    (a) (1+i)^5
    _______

    1 - i

    So I categorised each as z1 and z2, worked out the modulus and argument of each, ( |z1| = root 2, |z2| = root 2, arg z1 = pie/4, arg z2 = - pie/4)

    Carried out de moivre's theorem and got as far as the step

    4cis(5pie/4)
    __________

    cis(-pie/4)
    Think [tex]\frac{x}{y}[tex] as xy^{-1} and subtract those two arguments.

    It would be much appreciated if anyone could help me finish it off or correct me if I've gone wrong somewhere along the line. Cheers
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  3. #3
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    Do you think you could talk me trough it please. It's just that what you said there doesn't make sense to me as it's not a method I'm familiar with, are you American by chance, that might explain it :P
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  4. #4
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    Quote Originally Posted by scott.b89 View Post
    Do you think you could talk me trough it please. It's just that what you said there doesn't make sense to me as it's not a method I'm familiar with, are you American by chance, that might explain it :P
    What method are you familiar with? If x= r cis(\theta) and y= R cis(\phi), what is xy?

    If x= r cis(\theta) and y= R cis(\phi), what is \frac{x}{y}?
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  5. #5
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    What method are you familiar with? If and , what is ?

    If and , what is ?

    I've still never seen it expressed in that way before :P but...

    x = 4cis(5pie/4)

    y = cis(-pie/4)

    I think?
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  6. #6
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    Because \frac{1}{{1 - i}} = \frac{{1 + i}}{2} we get \frac{{(1 + i)^5}}{{1 - i}} = \frac{{\left( {1 + i} \right)^6 }}{2}.

    We get \left( {1 + i} \right)^6  = \left( {\sqrt 2 } \right)^6 cis\left( {\frac{{6\pi }}{4}} \right).

    Now you can finish.
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  7. #7
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    Quote Originally Posted by scott.b89 View Post
    What method are you familiar with? If and , what is ?

    If and , what is ?

    I've still never seen it expressed in that way before :P but...

    x = 4cis(5pie/4)

    y = cis(-pie/4)

    I think?
    Not quite what I asked but okay, assuming that x and y are those, please answer my question.

    If you do not know how to multiply or divide numbers in that form, how did you get the expression for the power?
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