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Math

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If A=<2,1> and B=<-1,1>, find the magnitude and direction angle for
-2A +3B


Find all specified roots: Cube roots of 8i



A parallelogram has sides of length 32.7cm and 20.2cm. If the longer diagonal has a length of 35.1cm, what is the angle opposite this diagonal?

  • Math -

    I assume <2,1> and <-1,1) are vectors
    then -2A + 3B
    = <-4,-2) + <-3,3) = <-7,1>
    magnitude = √(49+1) = √50 or 5√2
    direction: let the angle be Ø
    tanØ = -1/7
    Ø = appr 171.9° with the x-axis

  • Math -

    For the 2nd: are you familiar with D' Moivres Theorem?
    let me know, and I will go through with this one

    3rd:
    make a sketch and use the cosine law

    35.1^2 = 32.7^2 + 20.2^2 - 2(20.2)(32.7)cosØ
    ( see if you get Ø = 100.7° )

  • Math -

    Not really, can you go through it with me?

  • Math -

    2nd:

    let z = 0 + 8i
    = 8( 0 + i)

    we need 0+i in the form cosØ + isinØ
    we need Ø, so that cosØ = 0 and sinØ = 1
    mmmmhhh, Ø = 90 comes to mind

    so z = 8(cos90° + isin90°)
    so the cube root, and using De Moivre's
    z^(1/3) = 8^(1/3) (cos 30 + isin30)
    = 2( √3/2 + (1/2)i )
    = √3 + i

    our angle could also have been Ø = 450°
    then z^(1/3) = 2(cos 150 + isin150)
    = 2( -√3/2 + 1/2 i )
    = -√3 + i

    or Ø could have been 810
    then z^(1/3) = 2(cos 270 + isin270)
    = 2(0 + i(-1))
    = -2i ----> a surprise ? no ?

    or Ø could have been -270
    z^(1/3) = 2(cos -90 + isin -90)
    = 2(cos90 - isin90)
    = 2(0 - i) = -2i , mmmhhhh?

    check: is (-2i)^ = 8i ???
    LS = (-2i)^3
    = -8i^3
    = -8(i^2)(i)
    = -8(-1)i = 8i , YES

    so going to left by subtracting multiples of 360 will always yield -2i
    to the right, adding 360 to previous angles with always yield either √3 + i or -√3+i

    so cube root of 8i = -2i, √3+i , -√3+i

  • Math -

    how did you get appr 171.9° in the first question?

  • Math -

    Jane, you will have to be able to change a complex number from standard form to trig form

    in general a + bi
    = (√(a^2 + b^2) [cosØ + isinØ]
    where Ø is such that tanØ = b/a

    e.g. change 3 + 4i to trig form
    = √(3^2+4^2) (cosØ + isinØ), Ø = appr 53.13°
    = 5(cos 53.13 + isin53.13)

    De Moivre said
    of z = r(cosØ + isinØ)
    then z^n = r^n( cos (Ø/n) + i sin (Ø/n) )

    e.g. let do (3+4i)^2
    (since we could easily do this algebraically ...
    (3+4i)^2 = 9 + 24i + 16i^2
    = 9 + 24i - 16
    = -7 + 24i )

    using De Moivre:
    (3+4i)^2
    = 5^2( cos 2(53.13) + isin 2(53.13))
    = 25(cos 106.26 + isin 106.26)
    = -7 + 24i (I kept all the decimals in my calculator to get the exact answer.

    This is what I did in problem #2, you might want to print out this explanation and go over it slowly and try a few of your own.
    It is a great theorem

  • Math -

    How did I get 171.9 ?

    from tanØ = -1/7
    I know that the tangent is negative in II and IV by the CAST rule
    but <-7,1> ends up in quadrant II
    so I took tan-inverse of 1/7 to get 8.13° , had it been in quad I
    in quad II we have 180-8.13 = appr 171.9

  • typo - Math -

    In all that lengthy typing, I just noticed an error.
    When I stated De Moivre's Theorem, it should of course have been:

    then z^n = r^n( cos (nØ) + i sin (nØ) )

    I did use it correctly in my work.

  • Math -

    got it. thank you so much!

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