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Math Help - fraction with complex parts

  1. #1
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    fraction with complex parts

    Hi Im new to complex numbers and I can't figure out this problem.

    (j20)(4 + j0.2)/
    4 + j(20 + j0.2)

    I'm not sure how how to go about doing this to get an answer of 3.77 + j0.944

    Can any one help me?
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  2. #2
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    1. Imaginary unit is denoted by i, not j.
    2. Are you sure about parentheses? The value of the expression you typed is -1.2 + 40i.
    3. If 4 + j(20 + j0.2) is enclosed in parentheses, the answer is still 3.82 + 0.93i, which is different from yours.
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  3. #3
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    on the tutorial it shows some of the working.

    (j20)(4 + j0.2)/
    4 + j(20 + j0.2)

    = (j20)(4 + j0.2)/
    4 + j20.2

    = (j20)(4 + j0.2)(4-j20.2)/
    (4 + j20.2)(4 -j20.2)

    =(j20)(4 + j0.2)(4-j20.2)/
    16 + 20.2^2
    then not sure how to work the numerator out

    this is 4 steps, the spacings hasnt worked unfortunately.
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  4. #4
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    If you want to use the fixed-width font and consecutive spaces, you can use the [code]...[/code] tags. However, it is better to write the expression in a single line using parentheses.

    Is your expression \displaystyle\frac{(20i)(4 + 0.2i)}{4 + i(20 + 0.2i)}? Then it is not equal to \displaystyle\frac{(20i)(4 + 0.2i)}{4 + 20.2i}.
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  5. #5
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    the problem initally started as:

    1/ + 1/
    j15 2.4 + j0.41

    Where can I find info on using the maths tags?
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  6. #6
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    Sorry, I don't understand this notation. Please write expressions in a single line using parentheses.
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  7. #7
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    do you mean

    (1)/(j15) + (1)/(2.4 + j0.41)
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  8. #8
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    Hello, suzidoyle635!

    I assume the original problem was written in Klingon.
    No one on this planet writes like that.

    Besides, there are obviously some typos . . .


    \dfrac{j20(4 + j0.2)}{4 + j(20 + j0.2)}

    \text{Answer: }\:3.77 + j0.944

    \text{I am }guessing\text{ that the problem is: }\;\dfrac{20i(4+0.2i)}{4 + 20i + 0.2i}

    \text{This simplifies to: }\;\dfrac{\text{-}4 + 80i}{4+ 20.2i}


    \displaystyle \text{Rationalize: }\;\frac{\text{-}4+80i}{4+20.2i}\cdot\frac{4=20.2i}{4-20.2i} \;=\;\frac{\text{-}16 + 80.8i + 320i + 1616}{16+ 408.04}

    . . . . . . . . \displaystyle =\;\frac{1600 + 400.8i}{424.04} \;=\;\frac{1600}{424.04} + \frac{400.8}{424.04}i

    . . . . . . . . =\;3.773228941 + 0.94519385i

    . . . . . . . . \approx\;3.77 + 0.945i

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  9. #9
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    Yeah thank you very much for that, problem solved.
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