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Proving of complex number

  1. Nov 10, 2012 #1

    loy

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    1. The problem statement, all variables and given/known data
    Given that, x+yi=3/[2+cos θ+i(sin θ)] , prove that x^2+y^2=4x-3


    2. Relevant equations
    r^2=x^2+y^2


    3. The attempt at a solution
    since we know that x=r sinθ , y=r cosθ,
    i multiply r on the right hand side to equal 3r/[2r+x+yi], then multiply its comjugate (2r+x)-yi to equal (3rx+6r^2 -3ryi) / [(2r+x)^2 + y^2] . then i equal the imaginary part on left handside, yi= -3ryi/[(2r+x)^2 + y^2],
    (2r+x)^2 + y^2 = -3r,
    4r^2+x^2+ 4rx +y^2 = -3r, from the Relevant equations,
    5r^2+4rx= -3r ,divide both sides by r
    5r=-4x-3,
    r=-(4x+3)/5,
    x^2+y^2= [-(4x+3)/5]^2, until here, i dont know how to proceed the question... anyone can help me?
     
  2. jcsd
  3. Nov 10, 2012 #2

    HallsofIvy

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    We know that if θ is the angle the line from 0 to x+ yi makes with the positive real axis. Is that given? You don't mention it in the statement of the problem.

    If that is true then the right side is simply 3/(2+ x+ iy) and it would seem reasonable to multiply both sides by 2+ x+ iy giving (x+ iy)(2+ x+ iy)= 2x+ 2iy+ x^2+ ixy+ ixy- y^2= 3. That is NOT, in general, x^2+ y^2= 4x- 3.
     
  4. Nov 10, 2012 #3

    loy

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    not given , my lecturer just gave me the question as i've written above.
    So, what should I assume?
     
  5. Nov 10, 2012 #4

    SammyS

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    But you did assume that x=r sinθ and y=r cosθ , so what Halls said about θ is true.

    An alternate way to solve this problem, with no assumptions is to rationalize the denominator of [itex]\displaystyle \ \ \frac{3}{2+\cos(\theta)+i\sin(\theta)}\ \ [/itex] and identify x and y by equating real parts and equating imaginary parts.

    Then compare x2 + y2 with 4x-3 .
     
  6. Nov 10, 2012 #5

    loy

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    how to do equating the parts? they are cos and sin there.
     
  7. Nov 10, 2012 #6

    SammyS

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    Rationalize the denominator of [itex]\displaystyle \ \ \frac{3}{2+\cos(\theta)+i\sin(\theta)}\ \ [/itex] to determine what is the real part and what is the imaginary part.

    Since that's equal to x + iy, then you have x is equal to the real part, and y is equal to the imaginary part.
     
  8. Nov 11, 2012 #7

    loy

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    you mean multiply the conjugate of 2+cos(θ)+isin(θ) in order to find the x and y?
     
  9. Nov 11, 2012 #8

    SammyS

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    Just so there's no misinterpretation, I mean:

    Multiply [itex]\displaystyle \ \ \frac{3}{2+\cos(\theta)+i\sin(\theta)}\ \ [/itex] by [itex]\displaystyle \ \ \frac{2+\cos(\theta)-i\sin(\theta)}{2+\cos(\theta)-i\sin(\theta)}\ \ [/itex] in order to find the x and y .
     
  10. Nov 15, 2012 #9

    loy

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    thanks dude , i found the answer!
     
  11. Nov 17, 2012 #10
    hey

    I could not get the solution

    I multiply and divide the RHS with its conjugate 2+cos∅-isin∅ which arrives at
    6+3cos∅+isin∅ 6+3cos∅+isin∅
    -------------------- = ---------------------------
    (2+cos∅)^2 +sin^2∅ 4+4cos∅+cos^2∅+sin^2∅

    How to equate the LHS
     
  12. Nov 18, 2012 #11

    SammyS

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    Use [ code] [ /code] tags to write such fractions.

    Code (Text):

        6+3cosθ+isinθ                         6+3cosθ+isinθ
    --------------------          =          ---------------------------
    (2+cosθ)^2 +sin^2∅                      4+4cosθ+cos^2θ+sin^2θ
     
    Don't forget, sin2(θ) + cos2(θ) = 1 .
     
  13. Nov 18, 2012 #12
    yeah i know sin2(θ) + cos2(θ) = 1 .

    but even then i'm stuck, can u help me further
     
  14. Nov 18, 2012 #13
    It should be 6 + 3 Cosθ - i 3 Sinθ.
    3 is missing in i Sinθ


    Anyway, I kept getting 9 / (5+4Cosθ) for (x + iy)(x - iy)
     
    Last edited: Nov 18, 2012
  15. Nov 19, 2012 #14

    loy

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    the solution is like this
    3/[2+cos θ+i(sin θ)]
    = [3(2+Cosθ)- 3 Sinθ i] / [(2+cos θ)^2 + (sin θ)^2]

    So, from LHS, we get
    x=3(2+Cosθ)/ [(2+cos θ)^2 + (sin θ)^2] while
    y=3 Sinθ/ [(2+cos θ)^2 + (sin θ)^2]

    On LHS: x^2+y^2
    =[9(2+Cosθ)^2+9Sinθ^2]/ [(2+cos θ)^2 + (sin θ)^2]^2
    =9/[(2+cos θ)^2 + (sin θ)^2]
    since the denominator and numerator has a common factor (2+cos θ)^2 + (sin θ)^2

    On RHS: 4x-3
    = 4*{3(2+Cosθ)/ [(2+cos θ)^2 + (sin θ)^2]} - 3
    ={24 + 12cosθ-3[(2+cos θ)^2 + (sin θ)^2]}/[(2+cos θ)^2 + (sin θ)^2]
    ={24+12cosθ-12-12cosθ-3cos2(θ)-3sin2(θ)}/[(2+cos θ)^2 + (sin θ)^2]
    ={24+12cosθ-12-12cosθ-3*(sin2(θ) + cos2(θ))}/[(2+cos θ)^2 + (sin θ)^2]
    =9/[(2+cos θ)^2 + (sin θ)^2]

    which is equal to the LHS
    It is proven.
     
  16. Nov 19, 2012 #15
    Head bang.

    It did not strike that I can use L.H.S = R.H.S.

    I tried to get 4x-3 from x^2 + y^2.
    x^2+y^2 can be derived from (x + iy)(x-iy), then tried to simplify the x^2+y^2 to 4x-3.

    Thank you very much Loy! Good day
     
    Last edited: Nov 19, 2012
  17. Nov 21, 2012 #16

    loy

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    welcome dude !
     
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