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Math Help - Complex Numbers Help!

  1. #1
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    Complex Numbers Help!

    How do I find the argument for a value such as:

    (-2(square root3) + 6i)/(2i)

    This kind of example confuses me. Do you divide the 6i by 2i to simplify?
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  2. #2
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    Multiply the equation by 2i/2i and then simplify.
    The denominator will become -4.
    The final answer is 3+sqrt(3)i. The in between you should do.
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  3. #3
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    \frac{{ - 2\sqrt 3  + 6i}}{{2i}}\left( {\frac{{ - 2i}}{{ - 2i}}} \right) = \frac{{12 + 4\sqrt 3 i}}{4} = 3 + \sqrt 3 i<br />
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  4. #4
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    Oh yes, -2i being the complex conjugate. Thanks.
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  5. #5
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    Euler's Formula

    In the attached question, the second pair of brackets. Using the formuala I get:

    ((1/sqrt3) * (sqrt3/2)) - ((1/sqrt3) * (1/2))

    How do I continue the question. Please help, Thanks.

    Edit: I think I got it using the second bracket as i, and multiplying by the first set of brackets in the question. Is the answer D?
    Attached Thumbnails Attached Thumbnails Complex Numbers Help!-complex-numbers.jpg  
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  6. #6
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    the answer is actually A. i cant write the Latex too well though to give steps
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  7. #7
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    Okay, I see where I went wrong. Got a wrong sign.
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  8. #8
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    How do I do this example. I've tried Eular's Forumla on the bottom line but I can't get to the correct answer.
    Attached Thumbnails Attached Thumbnails Complex Numbers Help!-complex-numbers2.jpg  
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  9. #9
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    Quote Originally Posted by haku View Post
    How do I do this example. I've tried Eular's Forumla on the bottom line but I can't get to the correct answer.

    you should be working writing the numerator as ae^{i \theta}
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  10. #10
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    Okay, doing that gives me:

    (-e^(i2Pi/3)) / (4e^(iPi/3))

    How do I solve from here?
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  11. #11
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    Quote Originally Posted by haku View Post
    Okay, doing that gives me:

    (-e^(i2Pi/3)) / (4e^(iPi/3))

    How do I solve from here?
    you made a mistake, try finding the mod and arg again
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  12. #12
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    Okay, the modulus is 2, and the arg is 120 because it's in the second quadrant. So:

    ((-2e^(i2Pi/3))/(4e^(iPi/3))

    Is that correct?
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  13. #13
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    Quote Originally Posted by haku View Post
    Okay, the modulus is 2, and the arg is 120 because it's in the second quadrant. So:

    ((-2e^(i2Pi/3))/(4e^(iPi/3))

    Is that correct?
    i got the numerator as 2e^{i \frac{2 \pi}{3}}
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  14. #14
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    Yes, 120 degrees. That's what I attempted to type in. How does it simplify from then on? Do you subtract the imaginary numbers to give answer D?
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  15. #15
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    yes the answer is D.
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