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Math Help - Complex Numbers

  1. #1
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    Complex Numbers

    Hey guys, I'm totally stumped on this question. Any help will be appreciated.

    Question:

    For a real number \theta, define

     e^{i \theta} = \cos \theta + i \sin \theta

    Show that the identity  e^{i(\theta + k)} = e^{i \theta}e^{i k} is equivalent to the trigonometric rules:

     \cos (\theta + k) = \cos \theta \cos k - \sin \theta \sin k

     \sin (\theta + k) = \sin \theta \cos k + \cos \theta \sin k

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  2. #2
    Lord of certain Rings
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    Quote Originally Posted by WWTL@WHL View Post
    Hey guys, I'm totally stumped on this question. Any help will be appreciated.

    Question:

    For a real number \theta, define

     e^{i \theta} = \cos \theta + i \sin \theta

    Show that the identity  e^{i(\theta + k)} = e^{i \theta}e^{i k} is equivalent to the trigonometric rules:

     \cos (\theta + k) = \cos \theta \cos k - \sin \theta \sin k

     \sin (\theta + k) = \sin \theta \cos k + \cos \theta \sin k

    First assume  e^{i(\theta + k)} = e^{i \theta}e^{i k} and prove

     \cos (\theta + k) = \cos \theta \cos k - \sin \theta \sin k
     \sin (\theta + k) = \sin \theta \cos k + \cos \theta \sin k

    Idea:
    e^{i \theta} = \cos \theta + i \sin \theta
    And by given data we have,
     e^{i(\theta + k)} = e^{i \theta}e^{i k}
     e^{i \theta + i k} = \cos (\theta + k) + i \sin (\theta + k)
    Now,
    e^{i \theta}e^{i k}=(\cos \theta + i \sin \theta)(\cos k + i \sin k )
    Multiply term by term and don't forget to use i^2 = -1
    Then equate Real and Imaginary pats on both sides to get what you want.
    Now assume
     \cos (\theta + k) = \cos \theta \cos k - \sin \theta \sin k
     \sin (\theta + k) = \sin \theta \cos k + \cos \theta \sin k
    ad prove
     e^{i(\theta + k)} = e^{i \theta}e^{i k}

    Idea:
    This is simple. Write the LHS using the definition, then substitute the sine and cosine equations. Then factorize and show it's the exponential product!
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  3. #3
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    Terrific stuff, Isomorphism. Thank you so much!
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