Multiplying & Dividing Complex Numbers

Discussion in 'Geometry and Trigonometry' started by nycmathguy, Feb 19, 2022.

  1. nycmathguy

    nycmathguy

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    Section 6.6

    Can you do 48 and 50 in step by step fashion? Thank you very much.



    Screenshot_20220219-165404_Samsung Notes.jpg
     
    nycmathguy, Feb 19, 2022
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  2. nycmathguy

    MathLover1

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    48 .
    3i(1-sqrt(2)*i)
    =3i*1-3i*sqrt(2)*i
    =3i-3sqrt(2)*i^2
    =3i-3sqrt(2)*(-1)
    =3i+3sqrt(2)
    =3sqrt(2) + 3i


    50 .

    (1+sqrt(3)*i)/(6-3i).........rationalize

    ((6+3i)(1+sqrt(3)*i))/((6-3i)(6+3i))

    ((6+3i)(1+sqrt(3)*i))/(6^2-(3i)^2)

    (6+6sqrt(3)*i+3i+3i*sqrt(3)*i)/(36-(9(-1))

    ((3 + 6sqrt(3))*i - 3sqrt(3) + 6)/(36+9)

    (3(1 + 2sqrt(3)) i - sqrt(3) + 2)/45

    ((1 + 2sqrt(3)) i - sqrt(3) + 2)/15

    ((1 + 2sqrt(3)) i/15 - (sqrt(3) - 2)/15

    -(1/15) (sqrt(3) - 2) +(1/15)(1 + 2sqrt(3))*i
     
    MathLover1, Feb 19, 2022
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  3. nycmathguy

    nycmathguy

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    Thank you so much. We continue with Section 6.6 today.
     
    nycmathguy, Feb 20, 2022
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  4. nycmathguy

    nycmathguy

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    Each problem has three parts. You only did part (a). Can you do parts (b) & (c)?
     
    nycmathguy, Feb 25, 2022
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  5. nycmathguy

    MathLover1

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    48.

    b. product in trigonometric form

    3i(1-sqrt(2)*i)
    use 3i and (1-sqrt(2)*i),

    in trig. form they are

    z=3(cos(pi/2)+i*sin(pi/2))..........1)

    z=sqrt(3)(cos(-0.955316618)+i*sin(-0.955316618))..........2)

    multiply 1) and 2)

    3(cos(pi/2)+i*sin(pi/2))*sqrt(3)(cos(-0.955316618)+i*sin(-0.955316618))=4.24264+3* i

    c.

    product in standard form (I actually did it first time)

    3i(1-i*sqrt(2)*i)=3i-3i*sqrt(2)*i=3sqrt(2) + 3 i=4.24264+ 3i

    so, result is same

    you can do 50 same way
     
    MathLover1, Feb 26, 2022
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  6. nycmathguy

    nycmathguy

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    I will try 50 when time allows. My weekend was ruined. You know who messed it up.
     
    nycmathguy, Feb 26, 2022
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