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Math Help - I need help with complex numbers and real and imaginary.

  1. #1
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    Arrow I need help with complex numbers and real and imaginary.

    I'm in Acc Math 1, which is basically Honors Algebra 1. I have a quiz tomorrow and I really don't understand what we're learning because my teacher isn't very good.

    (3+5i)/2i

    (-1+7i)/(3-2i)

    (8+3i)/(3-i)

    Then, I have an idea about this answer, but I want to make sure I'm doing it correctly.
    Find real numbers x and y to make each of the following statements true.
    4x-4yi = 1-8i
    x=1/4 and y=2

    Finally, I know you have to FOIL this problem, but I still can't seem to get a logical answer.
    (3-8i)(-6-5i)

    I know there are a lot of problems, but even if someone helps me on just one I'll be very grateful. I'm lost, and I feel like I'm the only one in the class who is. Thanks so much!
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  2. #2
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    Quote Originally Posted by julienumbers View Post
    I'm in Acc Math 1, which is basically Honors Algebra 1. I have a quiz tomorrow and I really don't understand what we're learning because my teacher isn't very good.

    (3+5i)/2i

    (-1+7i)/(3-2i)

    (8+3i)/(3-i)

    Then, I have an idea about this answer, but I want to make sure I'm doing it correctly.
    Find real numbers x and y to make each of the following statements true.
    4x-4yi = 1-8i
    x=1/4 and y=2

    Finally, I know you have to FOIL this problem, but I still can't seem to get a logical answer.
    (3-8i)(-6-5i)

    I know there are a lot of problems, but even if someone helps me on just one I'll be very grateful. I'm lost, and I feel like I'm the only one in the class who is. Thanks so much!
    Are you asking to simplify these?

    1. \frac{3 + 5i}{2i} = \frac{i(3 + 5i)}{2i^2}

     = \frac{3i + 5i^2}{-2}

     = \frac{-5 + 3i}{-2}

     = \frac{5 - 3i}{2}.


    2. \frac{-1 + 7i}{3 - 2i} = \frac{(-1 + 7i)(3 + 2i)}{(3 - 2i)(3 + 2i)}

     = \frac{-3 - 2i + 21i + 14i^2}{9 + 6i - 6i - 4i^2}

     = \frac{-3 + 19i - 14}{9 + 4}

     = \frac{-17 + 19i}{13}.


    3. \frac{8 + 3i}{3 - i} = \frac{(8 + 3i)(3 + i)}{(3 - i)(3 + i)}

     = \frac{24 + 8i + 9i + 3i^2}{9 + 3i - 3i - i^2}

     = \frac{24 + 17i - 3}{9 + 1}

     = \frac{21 + 17i}{10}.


    4. 4x - 4iy = 1 - 8i

    Equating real and imaginary components gives

    4x = 1 and -4y = -8.

    Therefore x = \frac{1}{4} and y = 2.


    5. (3-8i)(-6-5i) = -18 -15i + 48i + 40i^2

     = -18 + 33i - 40

     = -58 + 33i.
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  3. #3
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    Quote Originally Posted by Prove It View Post
    Are you asking to simplify these?

    1. \frac{3 + 5i}{2i} = \frac{i(3 + 5i)}{2i^2}

     = \frac{3i + 5i^2}{-2}

     = \frac{-5 + 3i}{-2}

     = \frac{5 - 3i}{2}.


    2. \frac{-1 + 7i}{3 - 2i} = \frac{(-1 + 7i)(3 + 2i)}{(3 - 2i)(3 + 2i)}

     = \frac{-3 - 2i + 21i + 14i^2}{9 + 6i - 6i - 4i^2}

     = \frac{-3 + 19i - 14}{9 + 4}

     = \frac{-17 + 19i}{13}.


    3. \frac{8 + 3i}{3 - i} = \frac{(8 + 3i)(3 + i)}{(3 - i)(3 + i)}

     = \frac{24 + 8i + 9i + 3i^2}{9 + 3i - 3i - i^2}

     = \frac{24 + 17i - 3}{9 + 1}

     = \frac{21 + 17i}{10}.


    4. 4x - 4iy = 1 - 8i

    Equating real and imaginary components gives

    4x = 1 and -4y = -8.

    Therefore x = \frac{1}{4} and y = 2.


    5. (3-8i)(-6-5i) = -18 -15i + 48i + 40i^2

     = -18 + 33i - 40

     = -58 + 33i.

    Oh, my problem was the whole conjugate part. Thank you so much! I get it now. I just needed some help.
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