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FP1 Question - Attempt at your own risk watch

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    FP1 question on complex numbers. really annoying question

    4.

    z= -4+6i

    a) Calculate arg z, giving your answer in radians to 3 d.p

    The complex number w is given by w = A/(2-i), where A is a positive constant. Given that |w|= (root)20

    (b) Find w in the form a+ib, where a and b are constants,


    (c) calculate arg w/z
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    (Original post by zaini_attack)
    FP1 question on complex numbers. really annoying question

    4.

    z= -4+6i

    a) Calculate arg z, giving your answer in radians to 3 d.p

    The complex number w is given by w = A/(2-i), where A is a positive constant. Given that |w|= (root)20

    (b) Find w in the form a+ib, where a and b are constants,


    (c) calculate arg w/z
    for part a, you just use inverse tan of 6/4. and then add pi to it, as it is on the other side.
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    if nobody answers this by midnight then PM me and i will do it for you
    i cba doing it now because its pretty simple stuff and you could find the answer in any textbook
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    bump bump bump
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    fine well answer to first question is 2.15879...
    2.159 (3.d.p)
    method; get gfx calculator
    press math
    across twice to CPX
    down to "angle("
    press enter
    press -4+6i
    press enter
    truncate answer to 3dp with roundng
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    part C is exactly the same
    and for part B you need to remember how you get mod Z, draw an argand plane (or plain? hah) and then consider how the value of root20 could be made...
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    WOOOOOW. thanks alot. any help with the rest.
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    sorry but i cba doing B for you because it would take me more than about 3 seconds and im going to bed now.. ill do it for you tomorrow night if nobody has still answered
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    loooool. thanks alot anywayz
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    its fine FP1 is by far my fave module so far
    cba with stuff like C1 its just boring and i always drop a crap load of marks because i dont think about what im doing lol
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    w = A/(2-i) = A(2+i)/(2-i)(2+i) = (2A + Ai)/5, abs(w)=sqrt((2A/5)^2+(A/5)^2) = sqrt(5A^2/25) = sqrt(A^2/5), now recall that abs(w)=sqrt(20) then we must have that 20=A^2/5, A^2 = 100 => A = +/- 10, however, A>0 so A = 10
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    im pretty sure this is FP2 complex numbers
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    (Original post by blackdragonthegreat)
    im pretty sure this is FP2 complex numbers
    This is far too easy for fp2...
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    (Original post by blackdragonthegreat)
    im pretty sure this is FP2 complex numbers
    Definately FP1
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    For c I get:  \pi + \mathrm{arctan}(8)

    Anybody care to confirm this?
 
 
 
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Updated: November 14, 2008
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