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Thread: HSC 2018 MX2 Marathon

  1. #26
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    Re: HSC 2018 MX2 Marathon

    Quote Originally Posted by pikachu975 View Post
    So let z = x + 2i

    (x+2i)^2 = x^2 + 4xi - 4
    Since it is real then Im(z^2) = 0
    so 4x = 0
    x = 0

    Hence z = 2i
    thank you Pikachu!!

  2. #27
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    Re: HSC 2018 MX2 Marathon

    Quote Originally Posted by pikachu975 View Post
    So let z = x + 2i

    (x+2i)^2 = x^2 + 4xi - 4
    Since it is real then Im(z^2) = 0
    so 4x = 0
    x = 0

    Hence z = 2i
    how about if re(z) is 2Im(z) and z^2-4i is real

    I cant get the answer.. the answer is +- (2+i)

  3. #28
    Junior Member 1729's Avatar
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    Re: HSC 2018 MX2 Marathon

    Quote Originally Posted by sssona09 View Post
    how about if re(z) is 2Im(z) and z^2-4i is real

    I cant get the answer.. the answer is +- (2+i)
    Last edited by 1729; 8 Nov 2017 at 10:09 PM.
    sssona09 likes this.

  4. #29
    -insert title here- Paradoxica's Avatar
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    Re: HSC 2018 MX2 Marathon

    Prove the ellipses:



    where k is an arbitrary real number

    have the same area as the ellipse

    Last edited by Paradoxica; 20 Nov 2017 at 12:53 AM.
    If I am a conic section, then my e = ∞

    Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.

  5. #30
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    Re: HSC 2018 MX2 Marathon

    Quote Originally Posted by Paradoxica View Post
    Prove the ellipses:



    have the same area as the ellipse

    Continuing on, prove that the area of the ellipse defined by



    has area


  6. #31
    This too shall pass Sy123's Avatar
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    Re: HSC 2018 MX2 Marathon




  7. #32
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    Re: HSC 2018 MX2 Marathon

    Letting the value of the solution to x = cos(x) be r, the following area expressions are obtained through integration:

    A₁ = sin(r) - r²/2

    A₂ = 1 - sin(r) + r²/2

    A₂ - A₁ = 1 - 2sin(r) + r²

    r > sin(r) (proof is trivial and left as an exercise)

    - 2sin(r) > - 2r

    1 - 2sin(r) + r² > 1 - 2r + r² = (1-r)² > 0

    A₂ - A₁ > 0

    A₂ > A₁
    If I am a conic section, then my e = ∞

    Just so we don't have this discussion in the future, my definition of the natural numbers includes 0.

  8. #33
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    Re: HSC 2018 MX2 Marathon


  9. #34
    Senior Member sida1049's Avatar
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    Re: HSC 2018 MX2 Marathon

    Quote Originally Posted by Pakka View Post
    Change to polar form, apply de Moivre's theorem, done. For all of them.

    Bachelor of Science (Advanced Mathematics)/Bachelor of Arts III, USYD

  10. #35
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    Re: HSC 2018 MX2 Marathon

    Could show me some working pls. Sorry for the trouble

  11. #36
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    Re: HSC 2018 MX2 Marathon

    Never mind, I got it. Cheers, the advice made things easier.
    sida1049 likes this.

  12. #37
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    Re: HSC 2018 MX2 Marathon

    This one pls
    Capture 3.JPG

  13. #38
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    Re: HSC 2018 MX2 Marathon

    Quote Originally Posted by Pakka View Post
    This one pls
    Capture 3.JPG
    Realise the denominator and use De Moivre's at the end to get the n in 2ntheta
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  14. #39
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    Re: HSC 2018 MX2 Marathon

    Appreciate some help with this: Find the range of values of |z| and arg(z) for |z-4-4i|=2sqrt(2)

  15. #40
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    Re: HSC 2018 MX2 Marathon

    Quote Originally Posted by CapitalSwine View Post
    Appreciate some help with this: Find the range of values of |z| and arg(z) for |z-4-4i|=2sqrt(2)
    Sketch the locus of z and note that the modulus and argument of z use the join of the origin to a point on the circle.

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